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03Kinematics in two or three dimensions: vectors, projectile motion and relative velocity
You walk 30 m east across a car park, then 40 m north to the door. A friend who drove straight in asks how far you actually walked from the gate. You add the two numbers and say 70 m, and the answer is wrong by 20 m; the gate is 50 m away. Last week's arithmetic, where a step backwards was a minus sign, has quietly stopped working.
By the end of this section you can add two or three motions that point in different directions and get a number and an angle you can defend, split any launched object into a horizontal problem and a vertical problem that share nothing but the clock, and work out what a moving observer sees when they are moving too.
In 60 seconds
A quantity with a direction cannot be added with a plus sign, and the fix is the whole section: break every such quantity into components along fixed axes, do ordinary one dimensional arithmetic on each axis separately, and reassemble at the end.
Components of a vector
$$A_x = A\cos\theta,\qquad A_y = A\sin\theta$$
the angle is measured counterclockwise from the positive x axis
two observers are both moving and you need what one sees of the other
Three most common mistakes
Adding magnitudes. Two legs of 30 m and 40 m at a right angle give 50 m, not 70 m, and no amount of care with the arithmetic will rescue a sum that was set up as a sum of numbers.
Letting gravity into the horizontal column. The horizontal equation of a projectile has no acceleration term in it at all, so the horizontal position is a straight multiplication and nothing more.
Using the range formula when the ball lands lower than it was launched. That formula was derived for equal launch and landing heights; off a cliff or a roof it can be short by a third.
The published weights are Midterm 1 20%, Midterm 2 20%, quizzes 10%, homework 5%, final 25% and laboratory 20%. Components and the projectile split are the machinery the rest of the mechanics course is written in, so they reappear inside later questions rather than only as questions of their own.
How much time do you have?
10 minutes
The two moves that unlock almost every question: turn a magnitude and an angle into two signed numbers, and split a flying object into a horizontal line and a vertical line that share only the time.
card, c-vector-components, c-projectile-motion, formula card
45 minutes
Everything that turns into a number on an exam script: component addition, the four projectile equations, the height and range results with the condition attached, and the subscript rule for two moving observers.
card, c-vector-components, c-vector-kinematics, c-projectile-motion, c-projectile-problems, c-relative-velocity, faded ladder, practice B
full read
Adds the graphical picture that makes the component rules obvious rather than memorised, the difference between and that exam markers keep punishing, and the reason a moving frame changes the but not the physics.
hook, c-vectors-and-, c-vector-components, c-vector-kinematics, c-projectile-motion, c-projectile-problems, c-relative-velocity, contrast pair, faded ladder, exam example, practice A to D, self audit
By the end of this section
Add and subtract vectors graphically by the tail to tip and parallelogram constructions, and state the largest and smallest possible magnitude of a sum.
Resolve a vector into components along chosen axes, add several vectors by components, and recover the magnitude and the correct quadrant of the result.
Distinguish position, , average velocity, and average speed in two dimensions, and obtain velocity and acceleration from a by differentiation.
Separate projectile motion into a horizontal problem with zero acceleration and a vertical problem with acceleration of magnitude g, coupled only through the elapsed time.
Compute , maximum height, range and impact velocity for a launch at any angle, including launches that land above or below the launch height.
Transform a velocity from one to another using the subscript chain rule, and apply it to river crossing and wind drift problems.
Syllabus coverage
Kinematics in Two or Three Dimensions
Position, displacement, velocity and acceleration as vectors; the independence of the components; projectile motion and its problem solving method; relative velocity between two frames
Four concepts share this token, because the phrase covers both the general vector description of motion and the one special case, constant downward acceleration, that the whole week is built around.
covered
Vectors
Vectors and scalars, graphical addition and subtraction, multiplication by a scalar, components and
Placed first, because every kinematic statement in the rest of the section is a vector statement and there is no way to write one down before the arrows have rules.
covered
Three dimensional components
The third component and the unit vector k, carried in the notation but never needed for a numerical answer here
The week line says two or three dimensions, so the notation is written in a way that survives a third axis, but every worked problem is planar. Nothing in this term's assessment has been announced as three dimensional, so treat the third component as notation you can read rather than a technique you owe.
off_syllabus
The vector product
Multiplying two vectors together, in either the scalar or the vector sense
Not needed anywhere this week, since every operation here is adding vectors or multiplying a vector by an ordinary number. The products of two vectors arrive with the sections that need them, on work and on rotation; you are not responsible for them now.
deferred
Recall first
The three constant acceleration equations, on one axis
For motion along a single straight line with constant acceleration $a$: $v = v_0 + at$, then $x = x_0 + v_0 t + \tfrac{1}{2}at^{2}$, and the time free one, $v^{2} = v_0^{2} + 2a(x-x_0)$.
Every projectile problem in this section is two of these three equations written twice, once for the horizontal axis with $a=0$ and once for the vertical axis with $a=-g$. Nothing new is added to them; they are only used twice at a time.
Free fall near the ground
An object moving freely near the surface of the Earth, with air resistance neglected, has a downward acceleration of magnitude $g = 9.80\ \mathrm{m/s^{2}}$, the same for a stone and for a feather in a vacuum, and independent of how fast the object is already moving.
That last clause is what makes this whole section possible. Because the downward acceleration does not depend on the sideways speed, a ball that is also moving sideways falls exactly as fast as one that is not.
Right triangle trigonometry
In a right triangle with hypotenuse $h$ and an angle $\theta$, the side next to the angle is $h\cos\theta$, the side opposite it is $h\sin\theta$, and the two shorter sides satisfy $\text{adj}^{2}+\text{opp}^{2}=h^{2}$. Also $\tan\theta = \text{opp}/\text{adj}$.
Resolving a vector into components is exactly this triangle, drawn with the vector as the hypotenuse, so the whole of the component method is one line of school trigonometry applied carefully.
The quadratic formula, and what two roots mean
The solutions of $at^{2}+bt+c=0$ are $t = \dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. When the unknown is a time, both roots are mathematically valid and the physical situation decides which one to keep.
A projectile that lands at a different height from its launch gives a quadratic in the time, and the negative root is the instant before the throw at which the same parabola passed through the landing height. It is discarded on physical grounds, not because it is wrong.
Differentiating a polynomial in time
If $x(t) = At^{n}$ then $\dfrac{dx}{dt} = nAt^{n-1}$, and the derivative of a sum is the sum of the derivatives. A constant differentiates to zero.
When a position is given as a formula rather than as a list of numbers, velocity and acceleration are found by differentiating each component separately, and the components are polynomials in every problem here.
Units, significant figures and dimensional checking
Every reported quantity carries a unit; a product or quotient keeps as many significant figures as its weakest input; and every term in a correct equation has the same dimensions, so a candidate formula whose terms disagree can be rejected without any numbers.
The range and height results of this section are the first formulas of the course complicated enough to be misremembered, and the dimensional check is what catches a misremembering in ten seconds.
Try it yourself first (3 questions)
1§03.4 — acceleration at the top of a vertical throw●●○○○
Three short questions before the section starts, to find out which of last week's results are still to hand. Getting them wrong costs nothing and is not recorded anywhere; a miss simply tells you which recall above to read twice. A ball is thrown straight up and is watched at the single instant when it is momentarily at rest at the top of its flight.
Air resistance neglected, $g = 9.80\ \mathrm{m/s^{2}}$
Find
(a) What is the acceleration of the ball at the highest point of its flight?
Hint 1/4
The question is about acceleration, not about velocity, and the two are allowed to disagree; ask what the acceleration would have to be for the ball to be at rest and stay there.
Hint 2/4
Acceleration is the rate at which velocity changes, not the velocity itself. While the object is in free fall the acceleration has magnitude $g$ and points downward at every instant of the flight.
Hint 3/4
Substitute the data from the question: the ball is in free fall for the whole flight, up, at the top and down, so $a_y = -g = -9.80\ \mathrm{m/s^{2}}$ throughout, including at the instant when $v_y = 0$.
Hint 4/4
So the acceleration at the top is $9.80\ \mathrm{m/s^{2}}$ directed downward, the same as everywhere else on the flight.
Show solution
Argue from the definition rather than from a memorised list, because the whole trap is that the definition and the intuition point in opposite directions.
Independent check by contradiction: if the acceleration really were zero at the top, the velocity would stop changing there, and a ball with zero velocity and zero acceleration stays where it is. Balls do not hang in the air, so the acceleration is not zero.
Zero velocity and zero acceleration are separate claims about separate quantities, and a question that gives you one is often testing whether you will hand it the other for free.
2§03.2 — sides of a right triangle from an angle●○○○○
A straight ladder rests against a vertical wall on level ground. This is the same triangle that a vector and its two components make, drawn with a ladder instead of an arrow, so the arithmetic you use here is the arithmetic the next section will ask for.
Given
Ladder length $L = 6.20\ \mathrm{m}$
Angle to the ground $\theta = 62.0^{\circ}$
Find
(a) How high up the wall does the top of the ladder reach?
(b) How far is the foot of the ladder from the wall?
Hint 1/4
Draw the triangle and name which side you want before reaching for a function; the wall is the side opposite the angle and the ground is the side next to it.
Hint 2/4
For a right triangle with hypotenuse $L$ and angle $\theta$ at the base, the opposite side is $L\sin\theta$ and the adjacent side is $L\cos\theta$.
Hint 3/4
Substitute the given data, $L = 6.20\ \mathrm{m}$ and $\theta = 62.0^{\circ}$: the height is $6.20\sin 62.0^{\circ}$ and the base is $6.20\cos 62.0^{\circ}$.
Hint 4/4
So the ladder reaches $5.47\ \mathrm{m}$ up the wall and its foot is $2.91\ \mathrm{m}$ from it.
Show solution
Both sides come from the hypotenuse directly rather than one from the other, so an error in the first does not propagate into the second and the Pythagoras check at the end is genuinely independent.
The side opposite the angle
$$h = L\sin\theta = 6.20\sin 62.0^{\circ}$$
the wall is opposite the angle at the base, and sine is the ratio of the opposite side to the hypotenuse
Answer $$\boxed{h = 5.47\ \mathrm{m},\qquad b = 2.91\ \mathrm{m}}$$
Check
Independent check with Pythagoras, which was not used to get either answer: $5.474^{2}+2.911^{2} = 29.96 + 8.47 = 38.44$, and $6.20^{2} = 38.44$. The two sides do belong to this hypotenuse.
A ladder at 62 degrees is steep, so the tall side should be the bigger of the two, and it is; a sine and cosine that come out the wrong way round announce themselves as soon as you ask which side ought to be longer.
3§03.4 — a straight line run at constant acceleration●○○○○
A car pulls away from a stop line on a straight road and the acceleration stays the same for the whole six seconds. This is the vertical column of every projectile problem you are about to meet, with the numbers made friendly.
Two different questions are being asked, so two different equations are needed; sort out which of the constant acceleration equations has a distance in it and which has a final speed.
Hint 2/4
With $v_0 = 0$ the two equations reduce to $x = \tfrac{1}{2}at^{2}$ and $v = at$.
Hint 3/4
Substitute the given data, $a = 2.40\ \mathrm{m/s^{2}}$ and $t = 6.00\ \mathrm{s}$: $x = \tfrac{1}{2}(2.40)(6.00)^{2}$ and $v = (2.40)(6.00)$.
Hint 4/4
So the car covers $43.2\ \mathrm{m}$ and reaches $14.4\ \mathrm{m/s}$.
Show solution
Starting from rest kills the $v_0 t$ term, so the two equations become one term each and there is nothing left to get wrong except the factor of one half.
the velocity equation is linear in time, which is why the distance is not simply speed times time here
Answer $$\boxed{x = 43.2\ \mathrm{m},\qquad v = 14.4\ \mathrm{m/s}}$$
Check
Independent check using the average velocity: for constant acceleration from rest the average speed is half the final speed, $7.20\ \mathrm{m/s}$, and $7.20\times 6.00 = 43.2\ \mathrm{m}$, which agrees with the position equation.
Notice that the distance is not the final speed times the time; that product would be 86.4 m, exactly twice too big, and the factor of two is the whole content of the one half in the formula.
Notation
symbol
reads as
means
watch out
$\vec{A}$
the vector A
a quantity that needs a direction as well as a size: an arrow, drawn to scale, whose tail may be slid anywhere without changing what it is
an arrow over the letter is not decoration; the same letter without it means something different in the very next line
$A\ \text{or}\ |\vec{A}|$
the magnitude of A
the length of the arrow, always positive or zero, carrying the unit of the quantity
a magnitude is never negative, so an answer of minus 5 m for a distance means a sign was carried into a place where only a length belongs
$A_x,\ A_y$
the x component and the y component of A
two ordinary signed numbers, the shadows the arrow casts on the two axes, which together carry exactly as much information as the arrow
these are signed and can be negative, unlike the magnitude; the sign is what tells you the quadrant
$\hat{\imath},\ \hat{\jmath},\ \hat{k}$
i hat, j hat, k hat
vectors of length one, no units, pointing along the positive x, y and z axes; they exist only to say which direction a number belongs to
the hat means length one, not the same thing as an arrow; multiplying by a unit vector changes direction, never size
$\vec{r},\ \Delta\vec{r}$
the position vector, and the displacement
the arrow from the chosen origin to where the object is, and the change in that arrow between two instants
displacement is a difference of two positions and does not know about the path taken between them
$\vec{v}_{AB}$
the velocity of A relative to B
what an observer travelling with B measures for the motion of A
the order of the two subscripts is the whole content of the symbol; swapping them reverses the arrow
$\theta_0$
the
the angle between the initial velocity and the horizontal, measured counterclockwise from the positive x axis
a problem that measures the angle from the vertical instead swaps every sine with a cosine, so read the sketch before writing the components
$g$
the free fall constant
the positive number 9.80 m/s squared, the magnitude of the near the surface of the Earth
g itself is positive; the minus sign that makes the motion downward lives in the equation, written once, and is never doubled up by also making g negative
Conventions used here
Which way is positive.
Unless a problem says otherwise, the x axis is horizontal and positive to the right, the y axis is vertical and positive upward, and the origin sits at the launch point or the starting point of the motion. Every sign in every equation on the page is read against that choice, and the choice is written down before any number is.
A sign convention that changes between two lines of the same solution produces an answer that is wrong by a factor of minus one and looks entirely reasonable. Fixing it once, in writing, is what makes the minus signs checkable.
How angles are measured.
Angles are measured counterclockwise from the positive x axis, so that $A_x = A\cos\theta$ and $A_y = A\sin\theta$ hold in all four quadrants without a special case. When a problem states an angle some other way, for example above the horizontal or west of north, it is redrawn and converted to this convention before the components are written.
The two component formulas are only unconditionally true for this one convention. Every other convention needs its own bookkeeping, and mixing two of them inside one solution is the commonest way an otherwise correct component calculation ends up in the wrong quadrant.
The sign of the free fall acceleration.
The symbol $g$ always stands for the positive number $9.80\ \mathrm{m/s^{2}}$. With up positive, the vertical acceleration is written as $a_y = -g$, and the minus sign appears exactly once, in that line. These notes never write $g = -9.80\ \mathrm{m/s^{2}}$.
Students who let g carry the sign end up with two minus signs in the same term half the time, and the resulting projectile accelerates upward. Keeping the constant positive means there is only ever one place for the sign to be wrong.
What is neglected in a projectile problem.
Every projectile calculation in this section neglects air resistance and treats the free fall acceleration as constant in both magnitude and direction over the whole flight. These are stated as assumptions in each solution, not assumed silently, and the section says where they start to fail: a light ball, a long flight or a high speed.
The model is not a small technical detail, it is the reason the horizontal and vertical motions separate at all. With air resistance the horizontal speed is no longer constant and every result in this section stops being exact.
Relative velocity subscripts.
The symbol $\vec{v}_{AB}$ means the velocity of A as measured by B. Two such symbols combine when the inner subscripts match, $\vec{v}_{AC} = \vec{v}_{AB} + \vec{v}_{BC}$, and reversing the pair reverses the vector, $\vec{v}_{BA} = -\vec{v}_{AB}$.
The whole difficulty of relative velocity is bookkeeping, not physics. One fixed reading of the subscripts turns every problem in the topic into a check that the inner labels cancel.
How many digits an answer keeps.
Answers in these notes are rounded to three significant figures unless the data given deserve fewer, in which case the data win. Intermediate results carry one extra guard digit and are never rounded on the way through; rounding happens once, at the end. This section adds one clause to that: every numerical answer carries its unit, and every angle is quoted in degrees together with the direction it is measured from.
An angle without a stated reference direction is not an answer; 30 degrees from the horizontal and 30 degrees from the vertical describe two different launches. Naming the reference is part of the number.
3.1Vectors, and why two motions at an angle refuse to be added
Gives direction its own arithmetic, so two motions at an angle combine into one instead of being added as numbers.
Last week a direction was a plus or a minus sign. The car park at the top of this page is where that runs out.
Solvable with what we have
30 m east then 40 m east: the walk ends 70 m east of the gate, and the plus sign did all the work.
30 m east then 40 m west: the walk ends 10 m west of the gate, because west was a minus sign.
A car at 25 m/s that slows to 10 m/s in the same direction has changed speed by 15 m/s, and the subtraction was ordinary.
A stone thrown up at 12 m/s and caught coming down at 12 m/s has had its velocity change by 24 m/s, once up is called positive.
Two people pulling a sledge along the same rope in opposite directions: the sizes subtract, and the sign says who wins.
Not solvable yet
30 m east then 40 m north: no sign convention on one line can hold both of those directions at once.
A boat that swims at 2.3 m/s across a river while the river carries it at 1.2 m/s downstream: the two speeds are not aimed at each other in any way a plus or a minus can express.
A ball moving at 20 m/s east that a moment later moves at 20 m/s north: the speed never changed, and yet something clearly did.
An aeroplane pointing north at 165 km/h in a wind blowing east at 45 km/h: the plane does not travel at 210 km/h, and it does not travel at 120 km/h either.
A ball thrown at 20 m/s at 35 degrees above the ground: even writing down its starting motion needs two numbers, not one.
Keep the arithmetic and the walk becomes $30 + 40 = 70\ \mathrm{m}$ from the gate. A tape measure says $50\ \mathrm{m}$: out by 20 m, or 40 per cent.
Why it fails
Adding 30 and 40 answers a different question: the path walked, which really is 70 m. What was asked is how far the door is from the gate. Ordinary addition throws away the one thing that made the legs different, that they pointed different ways, so its answer cannot depend on the angle between them.
RuleRule: adding and subtracting vectors graphically
Conditions
both must be the same quantity in the same unit; a displacement and a velocity have no sum
an arrow may be slid anywhere on the page, but never turned
multiplying by a positive $c$ scales the length; a negative $c$ also reverses the direction
the magnitude of a sum lies between $\vert A-B\vert$ and $A+B$, reaching those ends only for antiparallel and parallel vectors
Slide the second arrow until its tail sits on the head of the first; the sum is the arrow from the very first tail to the very last head. The other order draws the other two sides of the same parallelogram and lands on the same corner, which is why order does not matter. Subtraction is the same construction, second arrow reversed.
The two legs are drawn tail to tip, and the is the arrow that closes the triangle. Its length is 50 m, not 70 m, and the missing 20 m is the whole reason vectors need their own arithmetic.
Looks like this, but is not
This is the rule working: two displacements of 30 m and 40 m pointing the same way give $70\ \mathrm{m}$; here the magnitudes really do add.
This looks like the same rule and is a different one: at right angles the same two give $50\ \mathrm{m}$, and facing each other $10\ \mathrm{m}$. Magnitudes add only for parallel arrows, so writing $|\vec{A}+\vec{B}| = A+B$ has quietly assumed an angle of zero. Note too that $|\vec{A}-\vec{B}|$ is also $50\ \mathrm{m}$ for the perpendicular pair: subtracting does not always shrink things.
The 30 m east and 40 m north walk that started this section
Find how far the door is from the gate, and in what direction, for the walk described at the top of the page.
Given
First leg: $30\ \mathrm{m}$ due east
Second leg: $40\ \mathrm{m}$ due north
The two legs are at right angles to each other
Find
The straight line distance from the gate to the door, and the direction of that line.
Solution
The tail to tip construction is chosen over the parallelogram here only because the two legs really were walked one after the other, so the drawing matches the story and is easier to check against it.
the arithmetic is exact here because 30, 40, 50 is a whole number triangle
Get its direction
$$\tan\theta = \frac{40}{30} = 1.333$$
the angle is taken at the gate, between the eastward leg and the resultant, so the northward leg is opposite and the eastward leg adjacent
$$\theta = \arctan(1.333) = 53.1^{\circ}$$
north of east, and the direction has to be named as well as the number, since 53.1 degrees on its own is not an answer
Answer $$\boxed{D = 50\ \mathrm{m}\ \text{at}\ 53.1^{\circ}\ \text{north of east}}$$
Check
Independent check on the size without Pythagoras: the resultant must be longer than either leg and shorter than their sum, so it must lie between 40 m and 70 m. Fifty sits in that window, while the naive answer of 70 m sits exactly on the forbidden edge, which is the edge reserved for two legs pointing the same way.
One construction, one square root and one arctangent. The whole of the rest of this section is spent making this three step process work when the angle is not 90 degrees.
The distance walked and the distance from the start are two different questions with two different answers, 70 m and 50 m, and both are correct answers to their own question. Read which one is being asked before choosing an arithmetic.
Change in velocity of a car that turns from east to north at a steady 20.0 m/s
A car is driving east at 20.0 m/s. Eight seconds later it is driving north at 20.0 m/s, and the speedometer read 20.0 m/s the whole time. Find the change in its velocity.
Given
Initial velocity $\vec{v}_i$: $20.0\ \mathrm{m/s}$ due east
Final velocity $\vec{v}_f$: $20.0\ \mathrm{m/s}$ due north
Elapsed time $\Delta t = 8.00\ \mathrm{s}$, not needed for this part
Find
The vector change in velocity, as a magnitude and a direction.
Solution
The two velocities are drawn tail to tail rather than tail to tip, because a difference is being asked for and the difference arrow runs from the head of the one being subtracted to the head of the other.
equal legs, so the difference bisects the right angle between north and west
$$\Delta\vec{v}\ \text{points north west}$$
which is the only direction that can both cancel the eastward motion and supply the northward motion
Answer $$\boxed{|\Delta\vec{v}| = 28.3\ \mathrm{m/s},\ \text{directed north west}}$$
Check
Independent check by the law of cosines, which never uses components: with two sides of 20.0 and an included angle of 90 degrees, the third side is $\sqrt{400+400-2(400)\cos 90^{\circ}} = \sqrt{800} = 28.3$, agreeing with the construction.
The speed was 20.0 m/s before and 20.0 m/s after, and the velocity still changed by more than the speed itself. That single fact is why the rest of this section talks about velocity vectors and almost never about speeds.
Checkpoint
§03.1 — reading the angle back out of a resultant●●○○○
Half a minute on the range of a resultant. A hiker walks one leg of 4.0 km and a second leg of 3.0 km, and afterwards a map shows that the finish is exactly 5.0 km from the start in a straight line.
Given
First leg: $4.0\ \mathrm{km}$
Second leg: $3.0\ \mathrm{km}$
Straight line distance from start to finish: $5.0\ \mathrm{km}$
Find
(a) What was the angle between the two legs?
Hint 1/4
You are being given the resultant and asked for the angle, which is the reverse of the usual direction of travel; ask which special angles make the resultant easy to predict.
Hint 2/4
The magnitude of a sum runs from $\vert A-B\vert$ when the two are opposite, through $\sqrt{A^{2}+B^{2}}$ when they are perpendicular, up to $A+B$ when they are parallel.
Hint 3/4
Substitute the data from the question, $A = 4.0$ km and $B = 3.0$ km: opposite gives 1.0 km, parallel gives 7.0 km, and perpendicular gives $\sqrt{16+9} = 5.0$ km.
Hint 4/4
So the two legs met at a right angle, which is 90 degrees.
Show solution
Testing the three special angles first is quicker than solving the general law of cosines, and one of them lands exactly on the given number.
Bracket the possible answers
$$|A-B| \le R \le A+B \;\Rightarrow\; 1.0\ \mathrm{km} \le R \le 7.0\ \mathrm{km}$$
any resultant outside that window would be impossible, so 5.0 km is at least allowed
the cross term has to vanish, and that pins the angle uniquely in the range from 0 to 180 degrees
Answer $$\boxed{\phi = 90^{\circ}}$$
Check
Independent check by elimination: at 0 degrees the resultant would be 7.0 km and at 180 degrees it would be 1.0 km, so neither of those can produce 5.0 km, and the general relation has only one solution in between.
⚠ Adding the magnitudes and calling the result a resultant
the two numbers are sitting there and the plus sign works on every other quantity in the course, so the hand does it before the head objects
the word change suggests before and after in that reading order, and the subtraction gets written in the order the words appeared rather than final minus initial
wrong$$\Delta\vec{v} = \vec{v}_i - \vec{v}_f$$
right$$\Delta\vec{v} = \vec{v}_f - \vec{v}_i$$
⚠ Turning an arrow while sliding it
the tail to tip construction involves moving the second arrow, and it is easy to also swing it round so that it meets the first one neatly
wrong$$\text{slide and rotate }\vec{B}\ \Rightarrow\ \text{a different vector}$$
right$$\text{slide }\vec{B}\ \text{only: length and direction both unchanged}$$
3.2Components: turning every arrow into two signed numbers
Turns each arrow into two ordinary signed numbers, so that adding vectors becomes adding two columns.
The drawing gave 50 m and 53.1 degrees because the angle happened to be a right angle. Three legs at awkward angles would need a ruler, a protractor and a lot of faith, so we replace the drawing with arithmetic.
RuleRule: resolving a vector and adding by components
Conditions
$\theta$ is measured counterclockwise from the positive $x$ axis, and with any other reference direction the sine and the cosine can swap places
the components are signed numbers while the magnitude is not, so a negative component is normal and a negative magnitude is an error
the arctangent a calculator returns lies between $-90^{\circ}$ and $+90^{\circ}$, so it can name only two of the four quadrants; the signs of the components settle the other two
components add axis by axis, and no angle appears anywhere in that line of the work
Every arrow is two numbers in disguise: how far it reaches along x, and how far it reaches along y. Once each arrow has been replaced by its pair, adding arrows is nothing but adding two columns of ordinary numbers, and the angle only reappears at the very end when the answer is turned back into an arrow.
Where the two component formulas come from
Drop a perpendicular from the head of $\vec{A}$ onto the $x$ axis. The arrow is now the hypotenuse of a right triangle whose horizontal side is $A\cos\theta$ and whose vertical side is $A\sin\theta$, which is the first line of the box.
Squaring and adding those two sides gives $A^{2}(\cos^{2}\theta+\sin^{2}\theta) = A^{2}$, which is the second line, and their ratio is $\tan\theta$, which is the third.
For the sum, walk along $\vec{A}$ and then along $\vec{B}$. The total sideways travel is $A_x+B_x$ whatever order the legs were walked in, and the same holds vertically, because the horizontal travel of one leg never changes the vertical travel of the other.
Resolving one arrow into two signed numbers. The x component is negative because the shadow falls on the negative side of the axis, and that sign is the only thing that will later tell the calculator which quadrant the vector was in.
Looks like this, but is not
This is the rule working: for $A = 34.0\ \mathrm{m}$ at $\theta = 145^{\circ}$, the component is $A_x = 34.0\cos 145^{\circ} = -27.9\ \mathrm{m}$, negative because the arrow reaches to the left.
This looks like the same rule and is a different one: the arrow also makes an angle of 35 degrees with the negative $x$ axis, and $34.0\cos 35^{\circ} = +27.9\ \mathrm{m}$. The number is the same size and the sign is wrong, so a resultant built from it will come out in the wrong quadrant and the error will not show up until the very last line. Cosine of the angle is only the $x$ component when the angle is the one this section defined.
Resultant of three legs at 30.0, 130.0 and 250.0 degrees
A drone flies three legs one after another. Find the single displacement from start to finish, as a magnitude and a direction.
Given
Leg 1: $25.0\ \mathrm{m}$ at $30.0^{\circ}$
Leg 2: $18.0\ \mathrm{m}$ at $130.0^{\circ}$
Leg 3: $12.0\ \mathrm{m}$ at $250.0^{\circ}$
All angles counterclockwise from the positive $x$ axis
Find
The magnitude and direction of the resultant displacement.
Solution
Components are chosen over three successive graphical constructions because the angles are not special: a drawing would have to be accurate to about a millimetre to give three significant figures, and the arithmetic gives them for free.
Resolve each leg, keeping the signs the cosines give
Independent check on the size: the three legs are 25.0, 18.0 and 12.0 m long, so the resultant cannot exceed 55.0 m, and because the second and third legs point broadly back against the first it should be well under 25.0 m. A result of 16.2 m sits where that reasoning says it must.
Three resolutions, two column sums, one square root and one arctangent: nine calculator entries for three significant figures, against a drawing that could not have delivered them at all.
Nothing in the middle of that calculation was a vector. The whole method is a short trip out of vector language into ordinary arithmetic and back again, and the only places an angle appears are the first line and the last.
The arctangent that lands in the wrong quadrant
A vector has components $A_x = -3.0$ and $A_y = -4.0$, both in metres. Find its magnitude and its direction.
Given
$A_x = -3.0\ \mathrm{m}$
$A_y = -4.0\ \mathrm{m}$
Find
The magnitude, and the direction measured counterclockwise from the positive $x$ axis.
Solution
The sketch is drawn before the calculator is touched, because the calculator is about to give an answer that is off by exactly 180 degrees and there is no way to detect that from the number alone.
Independent check by going backwards: $5.0\cos 233.1^{\circ} = -3.0$ and $5.0\sin 233.1^{\circ} = -4.0$, which reproduces the data. Doing the same with 53.1 degrees gives plus 3.0 and plus 4.0, so that candidate fails the check outright.
The habit worth taking from this: work out the magnitude, then look at the two signs and say out loud which quadrant the arrow is in, and only then read the calculator. The arctangent is a two to one map and it cannot do the quadrant for you.
Checkpoint
§03.2 — components of a vector in the third quadrant●●○○○
Half a minute on signs. A vector of magnitude 8.0 m points at 210 degrees, measured counterclockwise from the positive x axis, which puts it below the axis and to the left of the origin.
Given
Magnitude $A = 8.0\ \mathrm{m}$
Direction $\theta = 210^{\circ}$ counterclockwise from the positive $x$ axis
Find
(a) Which pair gives the two components?
Hint 1/4
Before any arithmetic, ask where in the plane 210 degrees points, and therefore what sign each of the two components has to carry.
Hint 2/4
With the angle measured counterclockwise from the positive $x$ axis, $A_x = A\cos\theta$ and $A_y = A\sin\theta$ in every quadrant, signs included.
Hint 3/4
Substitute the data from the question, $A = 8.0\ \mathrm{m}$ and $\theta = 210^{\circ}$: $\cos 210^{\circ} = -0.866$ and $\sin 210^{\circ} = -0.500$.
Hint 4/4
So the components are $-6.9\ \mathrm{m}$ along $x$ and $-4.0\ \mathrm{m}$ along $y$.
Show solution
The signs are decided from the sketch first and the magnitudes second, so that a calculator in the wrong angle mode cannot silently flip the answer into another quadrant.
Fix the quadrant by sketch
$$180^{\circ} < 210^{\circ} < 270^{\circ}$$
third quadrant, so both components must come out negative before any number is computed
Independent check by rebuilding the magnitude: $\sqrt{6.93^{2}+4.00^{2}} = \sqrt{48.0+16.0} = 8.0$, which returns the given length, so the pair belongs to this arrow.
⚠ Using the angle to the nearest axis instead of the angle from the positive x axis
problems often quote an angle above the horizontal or west of north, and that quoted angle gets fed straight into the cosine without being converted
Says that position, velocity and acceleration are arrows, and that each axis then runs its own one dimensional story.
Components turned adding arrows into adding columns. The same trick, applied to the arrows that describe motion, turns a two dimensional motion problem into two one dimensional ones.
DefinitionDefinition: the kinematic vectors and their components
Conditions
the axes are fixed once and do not turn while the motion is being described
differentiating a vector means differentiating each component, because the unit vectors do not change with time
average quantities are differences over an interval; instantaneous ones are derivatives at an instant, and the two agree only when the motion is uniform
average speed is the path length over the time and is a scalar; it is not the magnitude of the average velocity
The position of an object is an arrow from the origin to it. Velocity says how fast the tip of that arrow is moving and which way, and acceleration says how fast the velocity arrow itself is changing. Because the axes stand still, each of those statements is really two statements running side by side, one per axis, and neither ever consults the other.
Why differentiating a vector means differentiating its components
Write the position as $\vec{r}(t) = x(t)\hat{\imath}+y(t)\hat{\jmath}$ and form the difference over an interval: $\Delta\vec{r} = \Delta x\,\hat{\imath}+\Delta y\,\hat{\jmath}$, because the unit vectors are the same at both ends.
Divide by $\Delta t$ and let the interval shrink. The limit acts on each coefficient separately, so the derivative of the vector is the vector of the derivatives.
That is the whole content of the independence of the components: the operation that produces velocity from position never mixes the $x$ coefficient with the $y$ coefficient.
Average velocity points along the straight chord from P to Q, whatever the path did in between. Instantaneous velocity at P points along the tangent, and the two only agree when the path between them is straight.
Looks like this, but is not
This is acceleration: a car on a straight road whose speedometer climbs from 10 m/s to 20 m/s is accelerating, and both the everyday word and the definition agree.
This looks like no acceleration and is acceleration: a car going round a bend with the speedometer locked at 20 m/s. The magnitude of $\vec{v}$ is constant, but its direction is changing, so $\Delta\vec{v}$ is not zero and neither is $\vec{a}$. The everyday word tracks the speedometer; the definition tracks the velocity vector, and the two part company the moment the road bends.
Velocity and acceleration from a position given as a formula
A particle moves in the plane with position $\vec{r}(t) = (2.0+3.0t)\hat{\imath} + (5.0-1.5t^{2})\hat{\jmath}$, with $\vec{r}$ in metres and $t$ in seconds. Find the velocity and acceleration, and evaluate the velocity at $t = 2.0\ \mathrm{s}$.
Given
$x(t) = 2.0+3.0t$ metres
$y(t) = 5.0-1.5t^{2}$ metres
$t$ in seconds
Find
General expressions for $\vec{v}(t)$ and $\vec{a}(t)$, and the magnitude and direction of $\vec{v}$ at $t = 2.0\ \mathrm{s}$.
Solution
Each component is differentiated on its own rather than trying to differentiate a magnitude, because the magnitude involves a square root of a sum and its derivative would be a far uglier object with no extra information in it.
Differentiate component by component
$$v_x = \frac{dx}{dt} = 3.0\ \mathrm{m/s}$$
the x coefficient is linear in time, so its rate is a constant and the horizontal motion is uniform
$$v_y = \frac{dy}{dt} = -3.0t\ \mathrm{m/s}$$
the y coefficient is quadratic, so its rate grows in proportion to the time, and the minus says the particle drifts downward more and more quickly
Differentiate once more for the acceleration
$$a_x = 0,\qquad a_y = -3.0\ \mathrm{m/s^{2}}$$
constant in both components, so this particle behaves exactly like a projectile with a weaker gravity than the real one
Independent check by finite differences, which uses no calculus: between $t = 1.99$ and $t = 2.01$ the $y$ coordinate goes from $-0.94015$ to $-1.06015$ metres, a change of $-0.12000$ m in $0.0200$ s, giving $-6.00$ m/s, which is the derivative the calculus produced.
A constant acceleration with zero horizontal component and a steady downward vertical one is not a special case invented for this example; it is exactly the situation the next two concepts are about.
Average acceleration of the car that turned from east to north
The car from earlier in this section changed from 20.0 m/s east to 20.0 m/s north over 8.00 s, with the speedometer reading 20.0 m/s throughout. Find its .
Given
$\vec{v}_i = 20.0\ \mathrm{m/s}$ due east
$\vec{v}_f = 20.0\ \mathrm{m/s}$ due north
$\Delta t = 8.00\ \mathrm{s}$
Find
The magnitude and direction of the average acceleration.
Solution
The change in velocity is taken as a vector and only then divided by the time, rather than differencing the two speeds, because differencing the speeds gives zero and zero is not the answer.
dividing a vector by a positive scalar scales the length and leaves the direction untouched
$$\text{direction: north west, the same as } \Delta\vec{v}$$
which is why the average acceleration points nowhere near either of the two velocities it connects
Answer $$\boxed{\bar{a} = 3.54\ \mathrm{m/s^{2}},\ \text{directed north west}}$$
Check
Order of magnitude check: 3.54 m/s squared is a little over a third of the free fall value 9.80, which is the sort of sideways push a passenger notices firmly but survives comfortably in a car making a brisk turn over eight seconds. A value of, say, 35 m/s squared would have meant an arithmetic slip.
Constant speed and zero acceleration are two different statements, and only the second one is about the vector. Any question that says the speed did not change is not telling you the acceleration was zero.
Checkpoint
§03.3 — constant speed on a bend●●○○○
Half a minute on the difference between speed and velocity. A bus drives round a long curved exit ramp and the driver keeps the speedometer pinned at exactly 50 km/h for the whole ramp.
Given
Speed constant at $50\ \mathrm{km/h}$ throughout
The direction of travel changes continuously along the ramp
Claim: because the speed does not change, the acceleration of the bus is zero
Find
(a) True or false, with one sentence of justification.
Hint 1/4
Ask which quantity the definition of acceleration is built from: the number on the speedometer, or the velocity vector with its direction attached.
Hint 2/4
Acceleration is the rate of change of the velocity vector, and a vector changes if either its length or its direction changes.
Hint 3/4
Apply that to the data in the claim: the length of $\vec{v}$ is fixed at 50 km/h, but the direction is turning the whole way round the ramp, so $\Delta\vec{v}$ is not the zero vector.
Hint 4/4
So the claim is false: the bus is accelerating throughout the ramp even though the speedometer never moves.
Show solution
The argument is made with a concrete pair of velocities rather than in words alone, because the vector subtraction is the part that people find hard to believe until they see it come out non zero.
a non zero change over a finite time is a non zero average acceleration, and shrinking the interval does not rescue it
Answer $$\boxed{\text{False: the acceleration is not zero}}$$
Check
Independent check from the passengers: on a curved ramp at steady speed you are still pushed sideways in your seat, and being pushed is the everyday signature of a velocity that is changing.
⚠ Treating constant speed as constant velocity
in one dimension the two really were the same thing, and last week's habits carry over silently into a plane
3.4Projectile motion: two columns that share only the clock
Splits a flying object into a steady horizontal motion and a free fall, joined by nothing but the elapsed time.
We now have a vector acceleration and a rule that each component runs its own one dimensional story. Feed in the one acceleration that everything near the ground shares, and the whole topic falls out.
TheoremModel: projectile motion with constant downward acceleration
Conditions
air resistance is neglected, so nothing acts horizontally and the horizontal velocity never changes
the flight is short and low enough that the free fall acceleration is constant in size and direction, with $g = 9.80\ \mathrm{m/s^{2}}$
the axes are chosen with $x$ horizontal and $y$ vertically upward, and the origin at the launch point unless stated otherwise
the object is in flight: the model starts the instant it leaves the hand or the ramp and stops the instant it touches anything
The horizontal shadow of the projectile slides along at a fixed speed and would keep doing so for ever. The vertical shadow is a ball thrown straight up and falling back. The two shadows never speak to each other; the only thing they share is the reading on the clock, and that is the whole of the coupling.
How the two columns come out of the one dimensional theory
The acceleration vector is $\vec{a} = -g\hat{\jmath}$, so its components are $a_x = 0$ and $a_y = -g$, both constant.
Each component of the motion is therefore a one dimensional constant acceleration problem, and last week's three equations apply to each of them unchanged.
Putting $a = 0$ into the position equation collapses it to $x = x_0+v_{x0}t$; putting $a = -g$ into it gives the vertical line of the box. Nothing has been added to the one dimensional theory, only used twice.
Two balls leave the table edge at the same instant, one simply dropped and one pushed sideways at 6.0 m/s. The lines joining them at each time are horizontal, which is what the independence of the two components looks like when you photograph it.
Looks like this, but is not
This is the model applying: a stone thrown off a bridge, a ball kicked across a pitch, a coin flicked off a table. All of them are dense, slow and short lived enough that the air changes very little.
This looks like the same situation and is not: a badminton shuttlecock, a ping pong ball with spin, or a rifle bullet over a kilometre. In each of those the air does enough work over the flight that the horizontal speed visibly falls, so $v_x$ is no longer constant and the two columns stop being independent. The model is not slightly wrong there, it is answering a different question, and the honest move is to say so rather than to quote three significant figures.
A ball rolling off a 1.20 m table at 2.50 m/s
A ball leaves the edge of a table horizontally. Find the time it is in the air, how far from the table edge it lands, and the speed and direction with which it hits the floor.
$g = 9.80\ \mathrm{m/s^{2}}$, air resistance neglected
Origin at the table edge, $x$ horizontal, $y$ upward
Find
The time of flight, the horizontal distance, and the landing speed with its angle.
Solution
The vertical column is solved first because it is the only one that contains the height, and the height is the only piece of data that can fix the time; going at the horizontal column first would stall immediately with two unknowns in one equation.
with the origin at the table edge and up positive, the floor is at minus 1.20 m; the initial vertical velocity is zero because the launch was horizontal
the two components are perpendicular by construction, so the speed is the hypotenuse
$$\alpha = \arctan\!\left(\frac{4.850}{2.50}\right) = 62.7^{\circ}\ \text{below the horizontal}$$
the angle is quoted with its reference direction, since a bare 62.7 degrees would be ambiguous
Answer $$\boxed{t = 0.495\ \mathrm{s},\quad x = 1.24\ \mathrm{m},\quad v = 5.46\ \mathrm{m/s}\ \text{at}\ 62.7^{\circ}\ \text{below the horizontal}}$$
Check
Independent check on the landing speed using the time free equation, which never mentions $t$: $v_y^{2} = 2gh = 2(9.80)(1.20) = 23.52$, so $v_y = 4.850$ m/s, agreeing with the value found from the time. Order of magnitude: about half a second to fall from table height is exactly what dropping a pen off a desk feels like.
Two equations, one square root and one arctangent. Notice that the mass of the ball was never asked for and never used.
The horizontal launch speed did not appear anywhere in the time calculation. That is the general lesson: the time a projectile spends in the air is decided entirely by the vertical column.
Finding a launch speed from where the stone landed
A stone is thrown horizontally from the top of a cliff and is seen to strike the sea 32.0 m from the base of the cliff. Find the speed at which it was thrown and the velocity with which it hits the water.
Given
Cliff height $h = 45.0\ \mathrm{m}$
Horizontal distance to the impact point $x = 32.0\ \mathrm{m}$
Thrown horizontally, so $v_{y0} = 0$
$g = 9.80\ \mathrm{m/s^{2}}$, air resistance neglected
Find
The initial speed, and the impact speed with its angle.
Solution
The time is extracted from the height rather than from the horizontal distance, because the height is the piece of data that involves the acceleration and therefore the only one that can pin a time on its own.
Time from the fall
$$h = \tfrac{1}{2}gt^{2} \;\Rightarrow\; t = \sqrt{\frac{2(45.0)}{9.80}}$$
written with the height as a positive number and the sign carried by the geometry, to keep the square root clean
$$t = \sqrt{9.184} = 3.0305\ \mathrm{s}$$
a guard digit is kept because this time is about to be used twice
the components combine as a right triangle, as always
$$\alpha = \arctan\!\left(\frac{29.70}{10.56}\right) = 70.4^{\circ}\ \text{below the horizontal}$$
steeper than 45 degrees because the fall was long enough for the vertical speed to overtake the horizontal one
Answer $$\boxed{v_{x0} = 10.6\ \mathrm{m/s},\qquad v = 31.5\ \mathrm{m/s}\ \text{at}\ 70.4^{\circ}\ \text{below the horizontal}}$$
Check
Independent check on the vertical speed without the time: $v_y^{2} = 2gh = 2(9.80)(45.0) = 882$, giving $v_y = 29.70$ m/s, which is what the time based route gave. Order of magnitude: 10.6 m/s is a hard overarm throw, which is what a 45 m cliff and a 32 m landing point ought to imply.
Read the two answers together: a person can throw at about 10 m/s, and the stone arrives at 31.5 m/s. Almost all of the impact speed was supplied by the fall, not by the thrower.
Checkpoint
§03.4 — dropped and thrown at the same instant●●○○○
Half a minute on the independence of the two columns, in the form the exam uses it. Two identical balls leave the same table edge at the same instant: one is simply released, the other is pushed horizontally so that it leaves the edge at 15 m/s.
Given
Both balls start at the same height above a level floor
Ball A is released from rest; ball B leaves horizontally at $15\ \mathrm{m/s}$
Air resistance neglected
Find
(a) Which ball reaches the floor first?
Hint 1/4
Ask which of the two columns decides when the floor is reached, and whether anything in that column knows about the sideways push.
Hint 2/4
The vertical equation is $y = y_0+v_{y0}t-\tfrac{1}{2}gt^{2}$, and the horizontal launch speed does not appear anywhere in it.
Hint 3/4
Substitute the data from the question: both balls have $v_{y0} = 0$ and the same $y_0$, and both have $a_y = -g$, so both vertical equations are identical and give the same $t$.
Hint 4/4
So they land at the same instant, and the only difference is where.
Show solution
The comparison is made by writing both vertical equations and looking at what differs, rather than by computing two times, because the point is that the two equations are literally the same equation.
Write the vertical column for each ball
$$\text{A}:\; y = y_0-\tfrac{1}{2}gt^{2}$$
released from rest, so the initial vertical velocity term is absent
$$\text{B}:\; y = y_0-\tfrac{1}{2}gt^{2}$$
launched horizontally, so its initial vertical velocity is also zero and the equation is identical
Solve the shared equation once
$$0 = y_0-\tfrac{1}{2}gt^{2}\;\Rightarrow\; t = \sqrt{\frac{2y_0}{g}}$$
one equation, one time, and no symbol for the horizontal speed anywhere in it
Answer $$\boxed{t_A = t_B = \sqrt{2y_0/g}}$$
Check
Independent check by photograph: the figure in this concept shows the two balls at four instants, and the lines joining them are exactly horizontal at every one of them, which is the same statement made with a camera instead of algebra.
⚠ Letting the acceleration into the horizontal equation
the position equation was memorised as one formula with three terms, and it gets written out in full on both axes without asking what the acceleration is on each
wrong$$x = v_{x0}t+\tfrac{1}{2}gt^{2}$$
right$$x = v_{x0}t \qquad (a_x = 0)$$
⚠ Using the launch speed as the initial vertical velocity
the number given in the problem is a single speed, and it gets dropped into whichever column is being written first
Split the launch velocity once, and the rest is bookkeeping: the vertical part alone decides how high and how long, and the horizontal part turns that time into a distance. The range peaks at 45 degrees and takes the same value for two complementary angles, because doubling an angle and doubling its complement give sines that are equal.
Where the height, time and range results come from
At the top of the flight $v_y = 0$, so $v_{y0}-gt_{\text{top}} = 0$ and $t_{\text{top}} = v_{y0}/g$. Substituting into the vertical position equation gives $H = v_{y0}^{2}/2g$, which is the height result once $v_{y0}$ is written as $v_0\sin\theta_0$.
For a landing at the launch height, set $y = 0$ in $y = v_{y0}t-\tfrac{1}{2}gt^{2}$. The two roots are $t = 0$ and $t = 2v_{y0}/g$, and the second is the time of flight. Notice it is exactly twice $t_{\text{top}}$, and notice that this step is where the equal height assumption enters.
The range is then $R = v_{x0}T = (v_0\cos\theta_0)(2v_0\sin\theta_0/g)$, and the double angle identity $2\sin\theta\cos\theta = \sin 2\theta$ compresses it into the boxed form.
One launch, everything labelled: the initial velocity split into the two components that run the two columns, the highest point where only the vertical part has been used up, and the range measured along the ground.
Looks like this, but is not
This is the range formula applying: a ball kicked from the ground at 20.0 m/s and 35.0 degrees lands back on the same flat pitch, and $R = v_0^{2}\sin 2\theta_0/g = 38.4\ \mathrm{m}$ is exact.
This looks like the same problem and is not: the same throw made from a 12.0 m roof. The formula returns 28.6 m, the honest calculation returns 42.6 m, and the formula is short by a third. Nothing about the formula looks conditional when you write it down, which is precisely why it has to be written down with its condition attached: launch height equals landing height.
A ball kicked at 20.0 m/s and 35.0 degrees across level ground
Find the maximum height, the total time in the air, the horizontal range, and the speed of the ball at the highest point.
Given
Launch speed $v_0 = 20.0\ \mathrm{m/s}$
Launch angle $\theta_0 = 35.0^{\circ}$ above the horizontal
Launch and landing at the same height, on level ground
$g = 9.80\ \mathrm{m/s^{2}}$, air resistance neglected
Find
The maximum height, the time of flight, the range, and the speed at the apex.
Solution
Everything is built from the two components rather than quoted from the boxed results, so that the same working would survive unchanged if the ground were not level; the boxed formulas are then used at the end only as a check.
only the vertical component is zero at the top; the ball is still travelling forward at its original horizontal speed
Answer $$\boxed{H = 6.71\ \mathrm{m},\quad T = 2.34\ \mathrm{s},\quad R = 38.4\ \mathrm{m},\quad v_{\text{apex}} = 16.4\ \mathrm{m/s}}$$
Check
Independent check on the height using the time free equation, which was not used above: $0 = v_{y0}^{2}-2gH$ gives $H = 131.6/19.6 = 6.71$ m. Independent check on the range using the boxed formula: $R = (400)\sin 70.0^{\circ}/9.80 = 38.4$ m. Order of magnitude: a 38 m kick that stays up for 2.3 s and rises about two storeys is an ordinary long pass.
One split, four short lines, and the boxed formulas used only as a second opinion rather than as the method.
The speed at the highest point is not zero, and it is not small: it is 16.4 m/s, which is 82 per cent of the launch speed. Only the vertical part of the motion is spent at the apex.
The same throw from a 12.0 m roof, where the range formula fails
A ball is thrown from the roof of a building at 18.0 m/s, 30.0 degrees above the horizontal. Find how far from the base of the building it lands and how fast it is travelling when it does.
Given
Launch height $y_0 = 12.0\ \mathrm{m}$ above the ground
Launch speed $v_0 = 18.0\ \mathrm{m/s}$ at $\theta_0 = 30.0^{\circ}$ above the horizontal
$g = 9.80\ \mathrm{m/s^{2}}$, air resistance neglected
Origin on the ground at the base of the building, $y$ upward
Find
The horizontal distance from the base to the landing point, and the impact speed with its angle.
Solution
The vertical equation is solved as a quadratic rather than by doubling the time to the apex, because doubling assumes the ball comes back to the height it left, and here it does not; the quadratic costs one extra line and is right in both cases.
the negative root, minus 0.896 s, is the earlier instant at which the same parabola passed through ground level, before the throw happened, so it is discarded on physical grounds
combining the two components at the landing instant
$$\alpha = \arctan\!\left(\frac{17.78}{15.59}\right) = 48.8^{\circ}\ \text{below the horizontal}$$
steeper on the way down than the 30 degrees it went up at, because the vertical speed gained more than it started with
Answer $$\boxed{x = 42.6\ \mathrm{m},\qquad v = 23.6\ \mathrm{m/s}\ \text{at}\ 48.8^{\circ}\ \text{below the horizontal}}$$
Check
Independent check on the landing vertical speed without the time: $v_y^{2} = v_{y0}^{2}+2g y_0 = 81.0+235.2 = 316.2$, so $v_y = 17.78$ m/s, matching. And the check that matters most: the range formula would have given $R = (324)\sin 60.0^{\circ}/9.80 = 28.6$ m, which is 33 per cent short of the honest 42.6 m.
One quadratic instead of one doubling. That is the entire price of a method that works whether or not the ground is level.
Whenever the launch and the landing heights differ, the flight is no longer symmetric: the second half is longer and steeper than the first, so nothing that was derived by doubling the time to the apex survives.
Checkpoint
§03.5 — the other angle with the same range●●○○○
Half a minute on the double angle in the range formula. A projectile is launched from level ground at 30 degrees above the horizontal and lands a distance R away, with air resistance neglected.
Given
Launch and landing at the same height
Launch angle $30^{\circ}$, launch speed $v_0$, range $R$
The same launch speed $v_0$ is used again at a different angle
Find
(a) At which other launch angle does the same speed give the same range $R$?
Hint 1/4
The range depends on the angle only through one trigonometric factor, so ask which other angle makes that factor take the same value.
Hint 2/4
On level ground $R = v_0^{2}\sin 2\theta_0/g$, and the sine of an angle equals the sine of its supplement.
Hint 3/4
Substitute the data from the question, $\theta_0 = 30^{\circ}$: the factor is $\sin 60^{\circ}$, and the other angle whose double has the same sine satisfies $2\theta = 120^{\circ}$.
Hint 4/4
So the complementary angle, 60 degrees, gives the same range.
Show solution
The question is answered from the structure of the formula rather than by computing two ranges, because the structure also explains why there are exactly two answers and not three.
Isolate the angle dependence
$$R = \frac{v_0^{2}}{g}\sin 2\theta_0$$
everything except the sine factor is the same for both shots, so the ranges are equal exactly when the sines are
the supplement is the only other solution with the doubled angle between 0 and 180 degrees
$$\theta_2 = 60^{\circ} = 90^{\circ}-\theta_1$$
which is the complement of the first angle, and the reason 45 degrees is the maximum: it is its own complement
Answer $$\boxed{\theta_2 = 60^{\circ}}$$
Check
Independent check with numbers: taking $v_0 = 20.0$ m/s, the 30 degree shot gives $400\sin 60^{\circ}/9.80 = 35.3$ m and the 60 degree shot gives $400\sin 120^{\circ}/9.80 = 35.3$ m, the same to three figures.
⚠ Using the range formula when the landing height differs from the launch height
the formula is short, memorable and carries no visible warning, and the condition that produced it was three lines back in the derivation
wrong$$R = \frac{(18.0)^{2}\sin 60.0^{\circ}}{9.80} = 28.6\ \mathrm{m}\quad\text{(from a 12.0 m roof)}$$
right$$0 = 12.0+9.00t-4.90t^{2}\Rightarrow t = 2.733\ \mathrm{s},\; x = 42.6\ \mathrm{m}$$
⚠ Setting the speed to zero at the top of the flight
the phrase the ball stops rising gets compressed into the ball stops, and the vertical component is mistaken for the whole velocity
⚠ Taking the time of flight as twice the time to the apex when the ground is lower
the symmetry of the parabola is genuinely true for a level flight and it is a very useful shortcut there, so it gets carried into the cliff and roof problems where it silently fails
3.6Relative velocity: what a moving observer measures
Converts a velocity measured by one observer into the velocity a second, moving observer would measure.
Every velocity so far has been measured by someone standing still. A boat in a river and an aeroplane in a wind are measured by two observers at once, and the two answers differ.
RuleRule: combining velocities between frames
Conditions
the two frames move at constant velocity relative to each other and neither of them turns
the speeds involved are ordinary everyday speeds, far below the speed of light, where velocities simply add
$\vec{v}_{AB}$ always means the velocity of A as measured by B, and the reading never changes inside a solution
the chain only closes when the inner subscripts match: $\vec{v}_{AB}+\vec{v}_{BC}$ is meaningful and $\vec{v}_{AB}+\vec{v}_{CB}$ is not
Read the subscripts like a chain: the inner pair has to match and then cancels, and whatever labels are left on the outside name the velocity you have built. Reversing a pair reverses the arrow and leaves its length alone, which is the formal version of the obvious fact that if a train passes you at 30 m/s then you pass the train at 30 m/s the other way.
Why the subscripts chain
Let $\vec{r}_{AC}$ be the position of A measured from C, and $\vec{r}_{AB}$, $\vec{r}_{BC}$ likewise. Positions chain exactly as arrows do: $\vec{r}_{AC} = \vec{r}_{AB}+\vec{r}_{BC}$, because that is one tail to tip construction.
Differentiate with respect to time. Each term becomes the corresponding velocity, and the chain survives the differentiation unchanged.
Setting $A = C$ in the result gives $\vec{0} = \vec{v}_{AB}+\vec{v}_{BA}$, which is the reversal rule.
The velocity triangle, not a map. The boat points straight across at 2.30 m/s while the water carries the whole picture downstream at 1.20 m/s, and the ground velocity is the arrow that closes the triangle at 27.6 degrees off the crossing direction.
Looks like this, but is not
This is the rule working: a boat that does 2.30 m/s in still water is pointed straight across a river flowing at 1.20 m/s, and its velocity over the ground is the vector sum, $2.59\ \mathrm{m/s}$ at 27.6 degrees off the crossing direction.
This looks like the same sum and is not: adding the two numbers to get $3.50\ \mathrm{m/s}$, or subtracting them to get $1.10\ \mathrm{m/s}$. Neither is available, because the two velocities are at right angles and neither helps nor hinders the other. The sum of the speeds is only ever the speed of the sum when the two arrows point the same way.
A boat pointed straight across a 260 m river with a 1.20 m/s current
A boat is steered so that it points straight at the far bank. Find its velocity relative to the ground, how long the crossing takes, and how far downstream it lands.
Given
River width $260\ \mathrm{m}$
Boat speed relative to the water $v_{BW} = 2.30\ \mathrm{m/s}$, pointed straight across
Water speed relative to the ground $v_{WG} = 1.20\ \mathrm{m/s}$, downstream
Axes: $x$ downstream, $y$ across towards the far bank
Find
The ground velocity as a magnitude and a direction, the crossing time, and the downstream drift.
Solution
The crossing time is taken from the across component alone rather than from the ground speed and some diagonal distance, because the across component is the only part of the motion that reduces the remaining width of the river.
Build the ground velocity from the chain
$$\vec{v}_{BG} = \vec{v}_{BW}+\vec{v}_{WG}$$
the inner subscript W cancels, leaving the velocity of the boat relative to the ground, which is what an observer on the bank sees
$$\vec{v}_{BG} = (1.20,\ 2.30)\ \mathrm{m/s}$$
the current contributes only downstream and the boat only across, so the components do not interfere
the boat is carried downstream at the water's speed for the whole crossing, and its own effort adds nothing here
Answer $$\boxed{v_{BG} = 2.59\ \mathrm{m/s}\ \text{at}\ 27.6^{\circ}\ \text{downstream},\quad t = 113\ \mathrm{s},\quad d = 136\ \mathrm{m}}$$
Check
Independent check by similar triangles, which uses none of the arithmetic above: the drift over the width should equal the downstream velocity over the across velocity. Indeed $135.7/260 = 0.522$ and $1.20/2.30 = 0.522$, and both equal $\tan 27.6^{\circ}$.
One vector sum and two divisions. Notice how much of the work never happened: the ground speed of 2.59 m/s was not used to find the time.
The current does not slow the crossing at all when the boat points straight across. It only decides where on the far bank you arrive, which is why this is the fast way over and the wrong way to reach a particular jetty.
Aiming upstream so that the same boat lands directly opposite
The same boat and the same river. Now the crossing must end at the point directly opposite the start. Find the heading required and the time the crossing takes.
Given
River width $260\ \mathrm{m}$
Boat speed relative to the water $v_{BW} = 2.30\ \mathrm{m/s}$
Water speed relative to the ground $v_{WG} = 1.20\ \mathrm{m/s}$, downstream
Required: the ground velocity must have no downstream component
Find
The angle upstream of straight across at which the boat must be pointed, and the crossing time.
Solution
The condition is imposed on the ground velocity first and the heading deduced from it, rather than guessing a heading and checking where the boat ends up, because the requirement is a statement about the sum and not about either part.
slower than before, because less of the boat's speed is now aimed at the far bank
Answer $$\boxed{\theta = 31.4^{\circ}\ \text{upstream},\qquad t = 133\ \mathrm{s}}$$
Check
Independent check on the across component by a different route: $v_{BW}\cos\theta = 2.30\cos 31.4^{\circ} = 1.963$ m/s, agreeing with the Pythagorean value. And a sanity check on the direction of the difference: 133 s is longer than the 113 s of the straight across crossing, which is what you would expect from spending part of your engine on fighting the river.
Two crossings, two different questions: the fastest crossing and the crossing that lands you where you meant to go are not the same manoeuvre, and no boat can have both.
Checkpoint
§03.6 — does the current slow the crossing●●●○○
Half a minute on which component does which job. A swimmer sets off from one bank pointed exactly at the far bank and never changes heading, in a river whose current flows steadily downstream.
Given
The swimmer's velocity relative to the water is perpendicular to the banks
The current is parallel to the banks
Claim: the current makes the crossing take longer than it would in still water
Find
(a) True or false, with one sentence of justification.
Hint 1/4
Ask which component of the ground velocity actually reduces the distance still to be swum, and whether the current contributes anything to that component.
Hint 2/4
The across component of the ground velocity is what closes the gap, and velocities add component by component, so a purely downstream current adds nothing across.
Hint 3/4
Apply that to the data in the claim: the swimmer supplies the whole across component and the current supplies none of it, so the across component is the same as in still water.
Hint 4/4
So the claim is false: the crossing takes exactly as long, and only the landing point moves.
Show solution
The argument is made in components rather than with the resultant speed, because the resultant speed really is larger and quoting it is exactly the trap the question is built on.
downstream first, across second; the swimmer supplies the second entry and the current the first
$$(v_{SG})_y = v_{SW}$$
the across component contains no contribution from the current at all
Time from the across component
$$t = \frac{w}{(v_{SG})_y} = \frac{w}{v_{SW}}$$
the same expression as for still water, so the time cannot have changed
Answer $$\boxed{\text{False: the crossing time is unchanged}}$$
Check
Independent check with numbers from this section: 260 m at 2.30 m/s gives 113 s with or without the 1.20 m/s current, while the ground speed rises from 2.30 to 2.59 m/s. The faster ground speed is spent entirely on the extra downstream distance.
⚠ Adding the two speeds as ordinary numbers
both are given as single numbers with the same unit and the word combined suggests addition, so the vectors get lost on the way to the calculator
the two orders read almost identically out loud, and the wrong one still produces a plausible looking triangle
wrong$$\vec{v}_{BG} = \vec{v}_{BW}+\vec{v}_{GW}$$
right$$\vec{v}_{BG} = \vec{v}_{BW}+\vec{v}_{WG}$$
⚠ Using the ground speed to find the crossing time
the ground speed is the biggest number in the problem and it is the speed the boat really travels at, so it feels like the one that should divide the width
wrong$$t = \frac{260}{2.59} = 100\ \mathrm{s}$$
right$$t = \frac{260}{2.30} = 113\ \mathrm{s}$$
Adding any number of vectors, the order the steps have to happen in
Whenever two or more vectors of the same kind have to be combined and the angles are not right angles, or three significant figures are wanted.
Draw a rough sketch first
Not to scale, but with the four quadrants marked and every arrow roughly in the right one. This sketch is the only thing that will catch a quadrant error at the end, and it takes ten seconds.
Fix the axes and the angle convention in writing
State which way $x$ points and that angles run counterclockwise from it. Then convert every angle in the problem into that convention before touching a calculator; an angle given as above the horizontal or east of north is not yet in it.
Resolve each vector, letting the trig functions supply the signs
Write $A_x = A\cos\theta$ and $A_y = A\sin\theta$ with the converted angle. Do not insert minus signs by hand: if the angle is right, the cosine and sine are already negative where they should be.
Add the two columns separately
Total the $x$ entries, then total the $y$ entries. No angle and no square root belongs anywhere in this step; if one has appeared, something has gone in the wrong column.
Rebuild the magnitude
Apply Pythagoras to the two totals, never to the individual vectors. The result is positive by construction, so a negative magnitude means a slip.
Rebuild the direction, then fix the quadrant from the sketch
Take the arctangent of $R_y/R_x$, then compare it with the sketch. If the sketch says second or third quadrant and the calculator gave an angle between minus 90 and plus 90 degrees, add 180 degrees. Quote the angle with the direction it is measured from.
Where it goes wrong
Feeding in an angle measured from the wrong axis, which produces components of the right size and the wrong roles.
Adding the sizes of the components while ignoring their signs, which quietly turns a partial cancellation into a sum.
Accepting the arctangent without the sketch, which puts second and third quadrant answers exactly 180 degrees out.
Rounding each component to three figures before adding, which can cost the last digit of a resultant built from near cancelling terms.
Any projectile problem, in six steps that never change
Anything moving freely near the ground once it has been released: thrown, kicked, dropped, rolled off an edge or fired.
Draw the axes and put the origin somewhere convenient
Up positive, and the origin either at the launch point or on the ground, whichever makes more of the given numbers into zeros. Write the choice down; every sign afterwards is read against it.
Split the launch velocity
Compute $v_{x0} = v_0\cos\theta_0$ and $v_{y0} = v_0\sin\theta_0$ once, and put a box around $v_{x0}$: it is the same number for the entire flight.
Write the two columns side by side
Horizontal: $x = x_0+v_{x0}t$ and $v_x = v_{x0}$, with no acceleration term. Vertical: $y = y_0+v_{y0}t-\tfrac{1}{2}gt^{2}$ and $v_y = v_{y0}-gt$. Two columns on the page, physically separated, so that nothing migrates between them.
Find the time, from whichever column has enough information
The time is the only quantity the two columns share, so it is always the bridge. Usually the vertical column supplies it, because that is where the height and the acceleration live.
Take that time to the other column
Substitute it into the column you have not used yet, and read off what was asked. If two roots appeared, keep the one that happens after the launch.
Check the size, the sign and the units
Say what the answer is comparable to out loud. A range of tens of metres for a throw, a fall time of a fraction of a second from table height, an impact angle steeper than the launch angle whenever the landing is lower.
Where it goes wrong
Putting a $\tfrac{1}{2}gt^{2}$ term in the horizontal equation, where there is no acceleration at all.
Using the launch speed $v_0$ where the component $v_{y0}$ belongs, which is only harmless when the launch is vertical.
Doubling the time to the apex when the landing height differs from the launch height, which shortens the flight and the range together.
Setting the whole velocity to zero at the top instead of only its vertical component.
Relative velocity by subscript bookkeeping
Whenever two of the three objects in a problem are moving: a boat and a river, a plane and the air, two vehicles measured from each other.
Name the three labels
Usually the moving thing, the medium it moves through, and the ground. Write the three letters down before any velocity is assigned to them.
Tag every given velocity with an ordered pair
Each number in the problem is a $\vec{v}_{AB}$ for some specific pair, and deciding which pair is most of the work. The boat's own speed is relative to the water, not to the ground.
Write the chain you want and check the inner subscripts
Set up $\vec{v}_{AC} = \vec{v}_{AB}+\vec{v}_{BC}$ with the letters you actually have. If the inner pair does not match, reverse one of the given velocities using $\vec{v}_{BA} = -\vec{v}_{AB}$ until it does.
Resolve and add by components
Choose the axes along the natural directions of the problem, usually downstream and across, or east and north, and add the columns exactly as in the component method.
Answer the question that was actually asked
Crossing time comes from the across component alone; drift comes from the downstream component alone; and the heading to steer comes from setting an unwanted component to zero.
Sanity check the direction of the effect
Aiming into the current always costs time; a purely sideways current never costs time; and a tail wind always increases ground speed. If your answer disagrees with one of these, find out why before writing it down.
Where it goes wrong
Adding the two speeds arithmetically when the velocities are not parallel.
Dividing the river width by the ground speed rather than by the across component.
Reversing a subscript pair halfway through, which flips one arrow and still produces a plausible triangle.
Assuming the required heading is the same angle as the drift angle of the straight across crossing; they are 31.4 and 27.6 degrees for the same river, and neither substitutes for the other.
Average velocity of a jogger who runs east then north
A jogger runs 300 m due east in 100 s, then turns and runs 400 m due north in 200 s. Find the average velocity for the whole run.
Given
Leg 1: $300\ \mathrm{m}$ east in $100\ \mathrm{s}$
Leg 2: $400\ \mathrm{m}$ north in $200\ \mathrm{s}$
Find
The magnitude and direction of the average velocity over the whole run.
Solution
The whole run is treated as one interval rather than two, because average velocity depends only on where the runner ended up and how long it took, not on what happened in between.
Displacement, not distance
$$\Delta\vec{r} = (300,\ 400)\ \mathrm{m}$$
the two legs are perpendicular, so their components do not interfere and the totals are just the two leg lengths
Independent check by inequality: the average speed can never be smaller than the magnitude of the average velocity, because the path is never shorter than the straight line. Here 2.33 exceeds 1.67, as it must, and they would be equal only for a run in a single straight line.
Same runner, same stopwatch, two answers that differ by 40 per cent: 1.67 m/s for the average velocity and 2.33 m/s for the average speed, because the first divides the straight line from start to finish and the second divides the ground actually covered.
How to tell them apart
Ask whether the question can be answered by someone who saw only the start and the finish. If it can, it is a velocity question and you need the displacement; if the answer depends on the route taken, it is a speed or distance question. Also note the direction of the inequality: the average speed is always at least the magnitude of the average velocity, so a solution that produces the reverse has swapped them.
Scaffolding comes off
The common skeleton
Draw the axes, say which way is positive, and put the origin where it makes most of the given numbers zero.
Split the launch velocity into $v_{x0}$ and $v_{y0}$, and mark $v_{x0}$ as fixed for the whole flight.
Write the horizontal column with no acceleration and the vertical column with acceleration $-g$, side by side.
Find the elapsed time from whichever column has enough information, which is nearly always the vertical one.
Carry that time into the other column and read off the quantity that was asked for.
Check the magnitude, the sign and the units, and say what the answer is comparable to.
1 · fully worked
A ball thrown at 12.0 m/s and 40.0 degrees across level ground
Work the full skeleton once, with every reason written out. Find the range and the maximum height.
Given
$v_0 = 12.0\ \mathrm{m/s}$
$\theta_0 = 40.0^{\circ}$ above the horizontal
Launch and landing at the same height
$g = 9.80\ \mathrm{m/s^{2}}$
Find
The horizontal range and the maximum height.
Solution
The origin is put at the launch point rather than on the ground, because then $y_0 = 0$ and the landing condition is simply $y = 0$, which removes a constant from every line that follows.
Axes and origin
$$x\ \text{horizontal},\ y\ \text{up},\ \text{origin at the launch point}$$
this makes the landing condition $y = 0$ instead of $y = y_0$, which is one fewer symbol to carry
the time free vertical equation with $v_y = 0$ at the top, which avoids computing the apex time separately
Answer $$\boxed{R = 14.5\ \mathrm{m},\qquad H = 3.04\ \mathrm{m}}$$
Check
Independent check with the boxed range formula, which was not used in the working: $R = (144)\sin 80.0^{\circ}/9.80 = 14.47$ m. Order of magnitude: a 12 m/s throw carrying about 14 m and rising a bit over 3 m is an ordinary hard throw across a garden.
The range is about five times the maximum height for this angle. Keep that ratio in mind: an answer where the height comes out bigger than the range means an angle above 76 degrees, and if the problem said 40 degrees then something is wrong.
2 · you write the reasoning
Easier than the one above, because the launch is horizontal and there is no angle to split. A dart leaves a blowpipe horizontally at $14.0\ \mathrm{m/s}$ from a height of $1.60\ \mathrm{m}$ above a flat floor. How far from the blowpipe does it land? The steps are given; the reasons are not. Write the reason column yourself, then open the model reasons and compare.
$$v_{x0} = 14.0\ \mathrm{m/s},\qquad v_{y0} = 0$$
reasoning
The launch is horizontal, so the whole of the 14.0 m/s belongs to the horizontal column and the vertical column starts from rest. This is the only line where the word horizontal does any work, and it does all of it here.
$$-1.60 = -4.90t^{2}\;\Rightarrow\; t = \sqrt{\frac{2(1.60)}{9.80}} = 0.5714\ \mathrm{s}$$
reasoning
With the origin at the muzzle and up positive, the floor is at minus 1.60 m, and the initial vertical velocity term is absent because $v_{y0} = 0$. The vertical column is the only one containing the height, so it is the only one that can produce a time. The positive root is kept because the negative root is a time before the dart was fired.
The horizontal speed has not changed, so the distance is a single multiplication. The three significant figures come from the data, and the answer is comparable to the width of a room, which is what a blowpipe fired from chest height ought to give.
3 · find the buried error
Harder than the rung above, because the stone lands lower than it was launched, and the work has been done for you, badly. A stone is thrown from the top of a $15.0\ \mathrm{m}$ cliff at $16.0\ \mathrm{m/s}$, $25.0^{\circ}$ above the horizontal, and lands on the beach below. Exactly two of the four steps are wrong. Find both, and say what each one should have been.
Step 1. Origin at the base of the cliff, up positive. $v_{x0} = 16.0\cos 25.0^{\circ} = 14.5\ \mathrm{m/s}$ and $v_{y0} = 16.0\sin 25.0^{\circ} = 6.76\ \mathrm{m/s}$.
Step 2. Vertical column: $0 = 15.0+6.76t-4.90t^{2}$, so $4.90t^{2}-6.76t-15.0=0$ and the positive root is $t = 2.57\ \mathrm{s}$.
Step 3. Horizontal column: the stone accelerates while it flies, so $x = v_{x0}t+\tfrac{1}{2}gt^{2} = 14.5(2.57)+4.90(2.57)^{2} = 37.3+32.4 = 69.7\ \mathrm{m}$.
Step 4. At impact $v_y = 6.76-9.80(2.57) = -18.4\ \mathrm{m/s}$, so the speed is $v = 14.5+18.4 = 32.9\ \mathrm{m/s}$.
the two buried errors (2)
⚠ step 3
There is no acceleration in the horizontal direction, so the $\tfrac{1}{2}gt^{2}$ term does not belong in this column at all. The horizontal distance is the single product $x = v_{x0}t = 14.5(2.57) = 37.3\ \mathrm{m}$, and the extra 32.4 m is pure invention.
The position equation is memorised as one formula with three terms and then written out in full on both axes, because writing a formula down feels safer than leaving two thirds of it out. The step is also self reinforcing: the stone really is accelerating, so the sentence used to justify it sounds true.
The two components are perpendicular, so their magnitudes do not add. The impact speed is the hypotenuse, $v = \sqrt{14.5^{2}+18.4^{2}} = \sqrt{550} = 23.5\ \mathrm{m/s}$, which is smaller than the sum, as it always must be.
Both numbers are speeds in the same unit sitting next to each other on the page, and the word total invites addition. The wrong answer is also plausible in size, which is what keeps it alive through a check.
right
$v = \sqrt{(14.5)^{2}+(18.4)^{2}} = 23.5\ \mathrm{m/s}$, and independently $v^{2} = v_0^{2}+2gy_0 = 256+294 = 550$, so $v = 23.5\ \mathrm{m/s}$.
4 · the bare problem
§03.5 — a drive from a raised tee●●●●○
Nothing is filled in this time. A golf ball is struck from a tee that stands on a small mound above the fairway, and it lands on the level fairway below without hitting anything on the way.
Given
Launch speed $v_0 = 24.0\ \mathrm{m/s}$
Launch angle $\theta_0 = 52.0^{\circ}$ above the horizontal
The tee is $3.00\ \mathrm{m}$ above the fairway
$g = 9.80\ \mathrm{m/s^{2}}$, air resistance neglected
Find
(a) How long is the ball in the air?
(b) How far horizontally does it travel before landing?
(c) How fast is it moving when it lands?
Hint 1/4
Run the skeleton and notice which step changes: the launch and the landing are at different heights, so the symmetry shortcut for the flight time is not available.
Hint 2/4
Split the launch velocity, then solve $0 = y_0+v_{y0}t-\tfrac{1}{2}gt^{2}$ as a quadratic in $t$, keep the positive root, and carry it into $x = v_{x0}t$.
Hint 3/4
Substitute the data from the question, $v_0 = 24.0\ \mathrm{m/s}$, $\theta_0 = 52.0^{\circ}$ and $y_0 = 3.00\ \mathrm{m}$: $v_{x0} = 14.78\ \mathrm{m/s}$, $v_{y0} = 18.91\ \mathrm{m/s}$, and $4.90t^{2}-18.91t-3.00 = 0$.
Hint 4/4
So the ball is in the air for $4.01\ \mathrm{s}$, travels $59.3\ \mathrm{m}$, and lands at $25.2\ \mathrm{m/s}$.
Show solution
The origin is placed on the fairway rather than at the tee, so that the landing condition is $y = 0$ and the 3.00 m appears once, as the constant term of the quadratic, instead of in every line.
combining the two components at the landing instant
Answer $$\boxed{t = 4.01\ \mathrm{s},\quad x = 59.3\ \mathrm{m},\quad v = 25.2\ \mathrm{m/s}}$$
Check
Independent check on the landing speed without the time: $v^{2} = v_0^{2}+2gy_0 = 576+2(9.80)(3.00) = 634.8$, so $v = 25.2$ m/s, matching. Order of magnitude: a drive of about 60 m in four seconds is a modest but entirely ordinary golf shot, and the landing speed exceeds the launch speed by exactly the amount the 3.00 m drop supplies.
The extra 3.00 m of height bought about 2.4 m of extra distance and 1.2 m/s of extra landing speed. Small changes in the launch height do not change the answer much, but they do change the method, because the symmetry shortcut is gone the moment the two heights differ.
Full exam-style question
Recovering the launch from the range and the flight timeexam format
A projectile is launched from level ground and is observed to land 82.0 m away after 4.10 s in the air. Find its launch speed, its launch angle and the greatest height it reached. Air resistance is neglected.
Given
Horizontal range $R = 82.0\ \mathrm{m}$
Time of flight $T = 4.10\ \mathrm{s}$
Launch and landing at the same height, on level ground
$g = 9.80\ \mathrm{m/s^{2}}$
Find
The launch speed, the launch angle, and the maximum height.
Solution
This runs the standard method backwards, and it is worth noticing why that is possible at all: the range fixes the horizontal component on its own and the flight time fixes the vertical component on its own, precisely because the two columns are independent.
Independent check on the height by a route that never uses $v_{y0}$: for a level flight $H = gT^{2}/8 = (9.80)(16.81)/8 = 20.6$ m. Order of magnitude: 28.3 m/s is about 102 km/h, which is a hard struck ball rather than a thrown one, and 20.6 m is roughly a seven storey building, which is what a four second hang time implies.
Two divisions, one square root, one arctangent, and no simultaneous equations at any point, because the two columns were solved separately.
A launch angle of 45.1 degrees is a tenth of a degree off the angle of maximum range, so this was as long a shot as that speed could produce: at exactly 45.0 degrees the range would still have been 82.0 m to three figures. Near the maximum the range hardly responds to the angle, about 0.05 m per degree, while at 20 degrees the same one degree change is worth more than 2 m. That is why aiming errors are cheap near 45 degrees and expensive at shallow angles.
Practice
A · concept 4 questions
1§03.5 — the velocity at the top of the arc●●○○○
A classmate defends a wrong answer by saying that a thrown ball must stop for an instant at the top of its path, because it has finished going up and has not yet started coming down.
Given
A ball is launched at $18.0\ \mathrm{m/s}$ at $40.0^{\circ}$ above the horizontal
Claim: at the highest point of its flight the velocity of the ball is zero
Find
(a) True or false, with one sentence of justification.
Hint 1/4
The claim is about the velocity vector, which has two components; check both of them rather than the one the argument is about.
Hint 2/4
At the apex $v_y = 0$, but $v_x = v_0\cos\theta_0$ throughout the flight, and the speed is $\sqrt{v_x^{2}+v_y^{2}}$.
Hint 3/4
Substitute the data from the question, $v_0 = 18.0\ \mathrm{m/s}$ and $\theta_0 = 40.0^{\circ}$: at the apex the velocity is $18.0\cos 40.0^{\circ} = 13.8\ \mathrm{m/s}$, horizontal.
Hint 4/4
So the claim is false: the ball is still moving at 13.8 m/s at the top, all of it forwards.
Show solution
The two components are examined separately, because the claim is true of one of them and false of the other, and lumping them together is exactly the error being tested.
Check each component at the apex
$$v_y = 0$$
the apex is defined by the vertical velocity changing sign, so it passes through zero there
Independent check by picture: if the velocity really were zero at the apex, the ball would be momentarily at rest in mid air and the second half of the trajectory would be a vertical drop. Trajectories are not shaped like that.
Whenever a problem says something stops, ask which component it is talking about. In two dimensions almost nothing stops entirely.
2§03.1 — the range of possible sums for two equal vectors●●○○○
Two displacement vectors have exactly the same magnitude, 6.0 m each, but the angle between them is not stated and could be anything.
Given
$|\vec{A}| = 6.0\ \mathrm{m}$
$|\vec{B}| = 6.0\ \mathrm{m}$
The angle between them is unknown
Find
(a) Which statement about the magnitude of $\vec{A}+\vec{B}$ is correct?
Hint 1/4
Rather than picking a formula, ask what the two extreme arrangements of the arrows are and what each of them produces.
Hint 2/4
The magnitude of a sum runs from $\vert A-B\vert$ when the two are antiparallel to $A+B$ when they are parallel, and takes every value in between.
Hint 3/4
Substitute the data from the question, $A = B = 6.0\ \mathrm{m}$: the lower end is $\vert 6.0-6.0\vert = 0$ and the upper end is $6.0+6.0 = 12\ \mathrm{m}$.
Hint 4/4
So the magnitude of the sum can be anything from 0 to 12 m.
Show solution
The two extreme cases are checked first and the general formula second, because the extremes are where the answer lives and the formula only confirms that nothing outside them is reachable.
Independent check at a third angle: at 90 degrees the sum should be $6.0\sqrt{2} = 8.5$ m, which sits inside the claimed window and is neither endpoint, as expected.
3§03.3 — comparing average speed with average velocity●●●○○
A hiker walks a closed loop through the hills and arrives back at the car park. A classmate claims that on some trips the average speed can come out smaller than the magnitude of the average velocity.
Given
A trip starts at one point and ends at another, possibly the same one
Claim: for some trips, average speed is less than the magnitude of the average velocity
Find
(a) True or false, with one sentence of justification.
Hint 1/4
Both quantities are divided by the same total time, so the comparison is really a comparison of the two numerators.
Hint 2/4
Average speed is the path length over the time; the magnitude of the average velocity is the straight line displacement over the same time.
Hint 3/4
Apply that to the data in the claim: the path from start to finish can never be shorter than the straight line between them, and for the closed loop the displacement is zero while the path is not.
Hint 4/4
So the claim is false: the average speed is always at least the magnitude of the average velocity, and for a closed loop it is strictly greater.
Show solution
The two definitions are put over a common denominator so that the comparison reduces to a statement about lengths, which is a geometric fact rather than a kinematic one.
Independent check with the closed loop in the question: the hiker returns to the car park, so the displacement is zero and the magnitude of the average velocity is exactly zero, while the average speed is clearly not. The inequality holds in the strongest possible way there.
This inequality is a free check on any two dimensional kinematics answer: if your average speed comes out below your average velocity magnitude, one of the two is wrong and you know it without rechecking either.
4§03.6 — two vehicles seen from each other●●●○○
Two cars approach the same crossroads on perpendicular roads. One is heading due north and the other due east, and both speedometers read the same value.
Given
Car A: $90\ \mathrm{km/h}$ due north
Car B: $90\ \mathrm{km/h}$ due east
Both speeds measured relative to the ground
Find
(a) What is the speed of car A as measured by the driver of car B?
Hint 1/4
The question asks for a velocity in a frame that is itself moving, so build the relative velocity as a difference of the two ground velocities rather than comparing speedometer readings.
Hint 2/4
Use $\vec{v}_{AB} = \vec{v}_{AG}-\vec{v}_{BG}$, which is the chain rule with the ground as the middle label.
Hint 3/4
Substitute the data from the question, $\vec{v}_{AG} = (0,\ 90)$ and $\vec{v}_{BG} = (90,\ 0)$ in km/h with $x$ east: the difference is $(-90,\ 90)$ km/h.
Hint 4/4
So the speed of A relative to B is $\sqrt{90^{2}+90^{2}} = 127\ \mathrm{km/h}$.
Show solution
The ground is used as the middle label in the chain because both given velocities are already measured relative to it, so no velocity has to be reversed except the one that the subtraction reverses anyway.
the inner subscript G cancels, and the reversal rule turns the velocity of the ground relative to B into minus the velocity of B relative to the ground
the two entries are equal in size, so the magnitude is 90 times the square root of two
Answer $$\boxed{|\vec{v}_{AB}| = 127\ \mathrm{km/h}\ \text{towards the north west}}$$
Check
Independent check by symmetry: swapping the two cars gives $(90,\ -90)$, the same length in the opposite direction, which is what the reversal rule demands and is a genuine constraint the answer had to satisfy.
Equal speedometer readings do not mean the cars are at rest relative to each other. Only equal velocities, direction included, would do that.
B · computation 7 questions
1§03.2 — sum and difference of two vectors in component form●●○○○
Two vectors are given directly in component form, so no resolving is needed and the whole question is about the bookkeeping of signs.
(a) Find $\vec{A}+\vec{B}$, and give its magnitude and direction.
(b) Find $\vec{A}-\vec{B}$, and give its magnitude and direction.
Hint 1/4
Both parts are the same operation on two columns; decide before starting which sign each entry carries in each part.
Hint 2/4
Add or subtract component by component, then use $R = \sqrt{R_x^{2}+R_y^{2}}$ and $\tan\theta = R_y/R_x$, checking the quadrant from the signs.
Hint 3/4
Substitute the given components, $\vec{A} = (3.0,\ -4.0)$ and $\vec{B} = (-2.0,\ 6.0)$ metres: the sum is $(1.0,\ 2.0)$ and the difference is $(5.0,\ -10.0)$.
Hint 4/4
So the sum is $2.2\ \mathrm{m}$ at $63^{\circ}$ and the difference is $11\ \mathrm{m}$ at $-63^{\circ}$.
Show solution
The subtraction is written as the addition of the reversed vector rather than done in the head, because the second vector already has one negative entry and reversing it by hand is where the signs usually go wrong.
Independent check that the two answers belong together: adding the sum and the difference must return twice the first vector, and $(1.0,\ 2.0)+(5.0,\ -10.0) = (6.0,\ -8.0) = 2\vec{A}$, which it does.
Notice that the difference is five times longer than the sum. Subtraction is not a smaller operation than addition, and which one is bigger depends entirely on where the two arrows were pointing.
2§03.2 — resolving a vector in the second quadrant●●○○○
A survey marker is recorded as a distance and a bearing from a base station, and the recording has to be converted into the two coordinates the map uses.
Given
Magnitude $A = 34.0\ \mathrm{m}$
Direction $145^{\circ}$ counterclockwise from the positive $x$ axis
Find
(a) Find the two components.
(b) Rebuild the magnitude and the angle from your components as a check.
Hint 1/4
Before computing anything, say which quadrant 145 degrees is in and therefore what sign each component must carry.
Hint 2/4
Use $A_x = A\cos\theta$ and $A_y = A\sin\theta$ with the angle exactly as given, and let the trigonometric functions supply the signs.
Hint 3/4
Substitute the given data, $A = 34.0\ \mathrm{m}$ and $\theta = 145^{\circ}$: $\cos 145^{\circ} = -0.8192$ and $\sin 145^{\circ} = 0.5736$.
Hint 4/4
So the components are $-27.9\ \mathrm{m}$ and $+19.5\ \mathrm{m}$.
Show solution
The check is done by rebuilding rather than by re-resolving, because re-resolving would repeat the same operation and could repeat the same mistake.
Independent check on the relative sizes: 145 degrees is 35 degrees past the vertical, so the vector should lean more sideways than upward, and indeed the horizontal component is the larger of the two in size.
The rebuilding step is worth doing every time until the sign work is automatic; it costs one line and it catches both the sign errors and the quadrant errors at once.
3§03.3 — velocity and acceleration from a position formula●●●○○
A tracking system reports the position of a model boat on a lake as a formula in the time, and the report has to be turned into a velocity and an acceleration.
Given
$\vec{r}(t) = 4.0t\,\hat{\imath} + (3.0t-0.50t^{2})\hat{\jmath}$, in metres
$t$ in seconds
Find
(a) Find $\vec{v}(t)$ and $\vec{a}(t)$.
(b) Find the speed and the direction of travel at $t = 4.0\ \mathrm{s}$.
Hint 1/4
Do not try to differentiate a magnitude; decide first which quantities are components and differentiate those.
Hint 2/4
Differentiate each component separately, since $\vec{v} = (dx/dt)\hat{\imath}+(dy/dt)\hat{\jmath}$ and the unit vectors are constant.
Hint 3/4
Substitute the given position, $x = 4.0t$ and $y = 3.0t-0.50t^{2}$: the derivatives are $v_x = 4.0$ and $v_y = 3.0-1.0t$, and at $t = 4.0\ \mathrm{s}$ these are $4.0$ and $-1.0$ metres per second.
Hint 4/4
So the speed at $t = 4.0\ \mathrm{s}$ is $4.1\ \mathrm{m/s}$, directed $14^{\circ}$ below the positive x axis.
Show solution
Each component is handled on its own throughout, because the two coordinates are independent functions of time and combining them early would only make the differentiation harder.
Differentiate once
$$v_x = \frac{d}{dt}(4.0t) = 4.0\ \mathrm{m/s}$$
linear in time, so the horizontal motion is uniform
Independent check by finite difference, which uses no calculus: between $t = 3.9$ and $t = 4.1$ the $y$ coordinate goes from $4.095$ to $3.895$ m, a change of $-0.200$ m in $0.200$ s, giving $-1.00$ m/s. This matches the derivative exactly, as a symmetric difference always does for a quadratic.
The vertical component crossed zero at 3.0 s and the speed was at its smallest there, 4.0 m/s, which is a useful reading: in two dimensions the speed is smallest when the velocity is parallel to whichever axis has no acceleration.
4§03.4 — a ball leaving a bench horizontally●●○○○
A ball rolls along a laboratory bench and leaves the edge horizontally, landing on the floor. The bench height and the rolling speed have both been measured.
Given
Bench height $h = 0.750\ \mathrm{m}$
Speed at the edge $v_{x0} = 1.80\ \mathrm{m/s}$, horizontal
$g = 9.80\ \mathrm{m/s^{2}}$, air resistance neglected
Find
(a) How long is the ball in the air?
(b) How far from the bench does it land?
(c) With what speed does it strike the floor?
Hint 1/4
Two of the three parts need a time and the data contain only one quantity that can produce one; find which column that quantity lives in.
Hint 2/4
Vertical column with $v_{y0} = 0$: $h = \tfrac{1}{2}gt^{2}$. Horizontal column with no acceleration: $x = v_{x0}t$. Landing speed from the two components.
Hint 3/4
Substitute the given data, $h = 0.750\ \mathrm{m}$ and $v_{x0} = 1.80\ \mathrm{m/s}$: $t = \sqrt{2(0.750)/9.80}$, then $x = 1.80t$.
Hint 4/4
So the flight lasts $0.391\ \mathrm{s}$, the landing point is $0.704\ \mathrm{m}$ out, and the impact speed is $4.24\ \mathrm{m/s}$.
Show solution
The vertical column is solved first because it is the only one containing the bench height, and the height is the only datum that can fix a time on its own.
Time from the fall
$$h = \tfrac{1}{2}gt^{2}\;\Rightarrow\; t = \sqrt{\frac{2(0.750)}{9.80}} = 0.3912\ \mathrm{s}$$
the launch is horizontal, so the initial vertical velocity term is absent
the two perpendicular components combine as a hypotenuse
Answer $$\boxed{t = 0.391\ \mathrm{s},\quad x = 0.704\ \mathrm{m},\quad v = 4.24\ \mathrm{m/s}}$$
Check
Independent check on the vertical speed without the time: $v_y^{2} = 2gh = 2(9.80)(0.750) = 14.70$, giving $v_y = 3.834$ m/s, which agrees. Order of magnitude: about four tenths of a second and about seventy centimetres out is exactly what a ball rolling off a bench does in a laboratory.
The ball lands nearly as far out as the bench is high, which is a useful rule of thumb: for a horizontal launch the landing distance and the drop are comparable whenever the launch speed is a couple of metres per second.
5§03.5 — a steep launch across level ground●●●○○
A ball is launched from ground level across a flat field and lands on the same level. The launch is deliberately steep, so the flight is long and high.
Given
Launch speed $v_0 = 30.0\ \mathrm{m/s}$
Launch angle $\theta_0 = 60.0^{\circ}$ above the horizontal
Level ground, $g = 9.80\ \mathrm{m/s^{2}}$, air resistance neglected
Find
(a) Find the maximum height reached.
(b) Find the total time in the air.
(c) Find the horizontal range.
Hint 1/4
Split the launch velocity first and then ask which of the two components each of the three questions actually needs.
Hint 2/4
Vertical column gives $H = v_{y0}^{2}/2g$ and $T = 2v_{y0}/g$; the horizontal column then gives $R = v_{x0}T$.
Hint 3/4
Substitute the given data, $v_0 = 30.0\ \mathrm{m/s}$ and $\theta_0 = 60.0^{\circ}$: $v_{x0} = 15.00\ \mathrm{m/s}$ and $v_{y0} = 25.98\ \mathrm{m/s}$.
Hint 4/4
So the ball reaches $34.4\ \mathrm{m}$, stays up for $5.30\ \mathrm{s}$ and lands $79.5\ \mathrm{m}$ away.
Show solution
The three results are built from the components rather than quoted from the boxed formulas, so that the same lines would still be valid if the ground turned out not to be level; the formulas then serve as an independent check.
one multiplication with the unchanged horizontal speed
Answer $$\boxed{H = 34.4\ \mathrm{m},\quad T = 5.30\ \mathrm{s},\quad R = 79.5\ \mathrm{m}}$$
Check
Independent check with the range formula, unused above: $R = (900)\sin 120.0^{\circ}/9.80 = 79.5$ m. Order of magnitude: 30 m/s is about 108 km/h, a very hard hit, and a 34 m apex with a five second hang time is what a high, steep hit looks like.
Compare with a 30 degree launch at the same speed: the range would be the same 79.5 m, but the height would be 11.5 m instead of 34.4 m and the flight would last 3.06 s instead of 5.30 s. Equal ranges do not mean equal trajectories.
6§03.6 — a swimmer heading straight across a river●●●○○
A swimmer sets out from one bank aiming straight at the far bank and holds that heading the whole way, while the river carries her steadily downstream.
Given
River width $88.0\ \mathrm{m}$
Swimmer speed relative to the water $1.15\ \mathrm{m/s}$, aimed straight across
Current speed relative to the ground $0.680\ \mathrm{m/s}$, downstream
Find
(a) How long does the crossing take?
(b) How far downstream does she land?
(c) What is her speed and direction relative to the ground?
Hint 1/4
Three questions, and only one of them needs the ground velocity; decide for each part which component of the motion is doing the work.
Hint 2/4
The across component alone closes the gap, so $t = w/v_{SW}$; the downstream component alone produces the drift, $d = v_{WG}t$; the ground velocity is the vector sum of the two.
Hint 3/4
Substitute the given data, $w = 88.0\ \mathrm{m}$, $v_{SW} = 1.15\ \mathrm{m/s}$ and $v_{WG} = 0.680\ \mathrm{m/s}$: $t = 88.0/1.15$, then $d = 0.680t$.
Hint 4/4
So the crossing takes $76.5\ \mathrm{s}$, she lands $52.0\ \mathrm{m}$ downstream, and her ground velocity is $1.34\ \mathrm{m/s}$ at $30.6^{\circ}$ off the crossing direction.
Show solution
The time is taken from the across component before anything else is computed, because every other answer depends on it and none of them depends on the ground speed.
Independent check by similar triangles: the drift divided by the width should equal the tangent of the drift angle. Here $52.0/88.0 = 0.591$ and $\tan 30.6^{\circ} = 0.591$, so the geometry closes.
Fifty two metres of drift on an eighty eight metre river is more than half the width, which is worth knowing before you choose where to get in.
7§03.6 — an aircraft in a crosswind●●●●○
An aircraft is flown on a fixed heading while a steady wind blows across its path, and afterwards the pilot works out both what happened and what she should have done.
Given
Airspeed $165\ \mathrm{km/h}$, that is the speed of the plane relative to the air
Heading held due north
Wind $45.0\ \mathrm{km/h}$ blowing from the west, that is towards the east
Find
(a) Find the velocity of the aircraft relative to the ground.
(b) What heading and ground speed would have been needed to travel due north?
Hint 1/4
The two parts ask opposite questions: the first gives the heading and asks for the result, the second gives the result and asks for the heading.
Hint 2/4
Use $\vec{v}_{PG} = \vec{v}_{PA}+\vec{v}_{AG}$ for part one; for part two impose $(v_{PG})_{\text{east}} = 0$ and solve for the heading angle.
Hint 3/4
Substitute the given data, $\vec{v}_{PA} = 165\ \mathrm{km/h}$ north and $\vec{v}_{AG} = 45.0\ \mathrm{km/h}$ east: the sum is $(45.0,\ 165)$ km/h, and for part two $165\sin\theta = 45.0$.
Hint 4/4
So the plane actually travels at $171\ \mathrm{km/h}$, $15.3^{\circ}$ east of north, and to go due north it must be headed $15.8^{\circ}$ west of north at a ground speed of $159\ \mathrm{km/h}$.
Show solution
Part two is not a small adjustment of part one but a different construction: in the first the airspeed vector is given and the ground vector is built, and in the second the ground direction is given and the airspeed vector has to be tilted until the unwanted component cancels.
what is left of the airspeed after part of it has been spent cancelling the wind
Answer $$\boxed{171\ \mathrm{km/h}\ \text{at}\ 15.3^{\circ}\ \text{east of north};\qquad \text{head}\ 15.8^{\circ}\ \text{west of north at}\ 159\ \mathrm{km/h}}$$
Check
Independent check on part two by rebuilding the sum: $165$ km/h at $15.8^{\circ}$ west of north has components $(-44.9,\ 158.7)$, and adding the wind $(45.0,\ 0)$ leaves $(0.1,\ 158.7)$, which is due north to the accuracy of the rounding.
Two angles that differ by half a degree, 15.3 and 15.8 degrees, answer two completely different questions. Reading which one a problem wants is most of the work in this topic.
C · exam level 4 questions
1§03.4 — a horizontal throw from a building●●●○○
A stone is thrown horizontally from the flat roof of a building and lands on level ground below. The building height and the throwing speed are both known, and the landing point has to be predicted before anyone goes to measure it.
Given
Height of the roof above the ground: $20.0\ \mathrm{m}$
Launch velocity: $8.00\ \mathrm{m/s}$, horizontal
$g = 9.80\ \mathrm{m/s^{2}}$, air resistance neglected
Find
(a) How far from the base of the building does the stone land?
Hint 1/4
Only one of the two columns contains anything that can fix the time; find it before touching the horizontal distance.
Hint 2/4
Vertical column with $v_{y0} = 0$ gives $t = \sqrt{2h/g}$, and the horizontal column then gives $x = v_{x0}t$ with no acceleration term.
Hint 3/4
Substitute the data from the question, $h = 20.0\ \mathrm{m}$ and $v_{x0} = 8.00\ \mathrm{m/s}$: $t = \sqrt{40.0/9.80} = 2.02\ \mathrm{s}$, then $x = 8.00\times 2.02$.
Hint 4/4
So the stone lands $16.2\ \mathrm{m}$ from the base of the building.
Show solution
The time is taken from the fall rather than guessed from the horizontal motion, because the horizontal equation contains two unknowns and the vertical one contains only the time.
the horizontal speed is unchanged throughout, so this is a single multiplication
Answer $$\boxed{x = 16.2\ \mathrm{m}}$$
Check
Independent check by a rough estimate: a 20 m fall is about two seconds, since $t = \sqrt{2h/g}$ is close to $\sqrt{4}$, and 8 m/s for two seconds is 16 m. The exact answer had to be near that, and it is. Order of magnitude: a five storey building and a landing point about three car lengths out.
The 20.0 m of height bought 2.02 s of flight. Doubling the height would not double the distance, because the time only grows as the square root: a 40 m building would give 2.86 s and 22.9 m, not 32.3 m.
2§03.5 — an angled throw from a cliff top●●●●○
A ball is thrown from the edge of a sea cliff at an angle above the horizontal, and it lands on the beach at the foot of the cliff. This is the shape of question where a memorised formula and an honest calculation part company.
Given
Cliff height above the beach: $25.0\ \mathrm{m}$
Launch speed $v_0 = 15.0\ \mathrm{m/s}$ at $\theta_0 = 40.0^{\circ}$ above the horizontal
$g = 9.80\ \mathrm{m/s^{2}}$, air resistance neglected
Find
(a) How far from the base of the cliff does the ball land?
Hint 1/4
Check first whether the shortcut you are about to use assumes anything about the landing height, because this landing is not at the launch height.
Hint 2/4
Solve the vertical equation $0 = y_0+v_{y0}t-\tfrac{1}{2}gt^{2}$ as a quadratic in $t$, keep the positive root, and carry it into $x = v_{x0}t$.
Hint 3/4
Substitute the data from the question, $y_0 = 25.0\ \mathrm{m}$, $v_0 = 15.0\ \mathrm{m/s}$ and $\theta_0 = 40.0^{\circ}$: $v_{x0} = 11.49\ \mathrm{m/s}$, $v_{y0} = 9.64\ \mathrm{m/s}$, and $4.90t^{2}-9.64t-25.0 = 0$.
Hint 4/4
So the flight lasts $3.45\ \mathrm{s}$ and the ball lands $39.6\ \mathrm{m}$ from the base.
Show solution
The quadratic is used rather than the range formula because the range formula was derived by setting the landing height equal to the launch height, and here they differ by the whole 25.0 m.
the negative root belongs to the part of the parabola before the throw and is discarded
Horizontal column
$$x = (11.49)(3.448) = 39.6\ \mathrm{m}$$
one multiplication, with the horizontal speed unchanged
Answer $$\boxed{x = 39.6\ \mathrm{m}}$$
Check
Independent check on the flight time from the landing speed: $v_y^{2} = v_{y0}^{2}+2gy_0 = 92.97+490.0 = 583.0$, so $v_y = 24.14$ m/s downward, and $(24.14+9.642)/9.80 = 3.448$ s, which reproduces the time by a route that never solved the quadratic.
The level ground formula would have given 22.6 m for the same throw, which is 43 per cent short. The size of that gap is the reason the condition on the range formula is worth memorising along with the formula.
3§03.6 — the heading that cancels a crosswind●●●●○
A light aircraft has to make good a track due north between two airfields while a steady wind blows across the route. The pilot has to choose a heading before departure, not correct it afterwards.
Given
Airspeed $240\ \mathrm{km/h}$
Wind $65.0\ \mathrm{km/h}$ blowing from the west, that is towards the east
Required track over the ground: due north
Find
(a) Which heading and ground speed are needed?
Hint 1/4
The requirement is a statement about the sum of two velocities, not about either of them, so write the condition on the sum first.
Hint 2/4
Impose that the eastward component of $\vec{v}_{PG} = \vec{v}_{PA}+\vec{v}_{AG}$ is zero, which gives $v_{PA}\sin\theta = v_{AG}$ with $\theta$ measured from north.
Hint 3/4
Substitute the data from the question, $v_{PA} = 240\ \mathrm{km/h}$ and $v_{AG} = 65.0\ \mathrm{km/h}$: $\sin\theta = 65.0/240 = 0.2708$, and the northward component is $\sqrt{240^{2}-65.0^{2}}$.
Hint 4/4
So the aircraft must be headed $15.7^{\circ}$ west of north and will make good $231\ \mathrm{km/h}$.
Show solution
The condition is imposed on the eastward component of the sum rather than on the plane's own velocity, because it is the ground track that must be north and the plane itself must point somewhere else.
the northward component that remains after part of the airspeed has been spent on the wind
Answer $$\boxed{15.7^{\circ}\ \text{west of north},\qquad 231\ \mathrm{km/h}}$$
Check
Independent check by rebuilding: $240$ km/h at $15.7^{\circ}$ west of north has components $(-64.9,\ 231.1)$ km/h, and adding the wind $(65.0,\ 0)$ leaves $(0.1,\ 231.1)$, which is due north to within the rounding.
Flying into a pure crosswind costs speed even though the wind is not against you: 240 km/h of airspeed makes good only 231 km/h, and the missing 9 km/h went into standing still sideways.
4§03.5 — comparing two launches at the same speed●●●○○
Two identical balls are launched from the same spot on level ground with exactly the same speed, one at 30 degrees above the horizontal and the other at 60 degrees. Air resistance is neglected for both.
Given
Same launch speed $v_0$ for both
Launch angles $30^{\circ}$ and $60^{\circ}$ above the horizontal
Level ground, both landing at the launch height
Find
(a) Which statement about the two flights is correct?
Hint 1/4
Two quantities are being compared at once, so work out separately how each of them depends on the launch angle before comparing anything.
Hint 2/4
On level ground $R = v_0^{2}\sin 2\theta_0/g$ and $H = v_0^{2}\sin^{2}\theta_0/2g$, so the range depends on the doubled angle and the height on the angle itself.
Hint 3/4
Substitute the two angles from the question: the ranges use $\sin 60^{\circ}$ and $\sin 120^{\circ}$, which are equal, while the heights use $\sin^{2}30^{\circ} = 0.25$ and $\sin^{2}60^{\circ} = 0.75$, which are not.
Hint 4/4
So the two ranges are equal while the 60 degree launch rises three times as high.
Show solution
The comparison is done with ratios rather than with numbers, because no launch speed was given and the ratios do not need one.
Independent check with numbers, taking $v_0 = 30.0$ m/s: the 30 degree launch gives $R = 79.5$ m and $H = 11.5$ m; the 60 degree launch gives $R = 79.5$ m and $H = 34.4$ m. The ranges agree and the heights are in the ratio three to one.
Equal ranges do not mean equal trajectories, and a question that asks which ball lands first is asking about the time, where the two differ by a factor of 1.73.
D · interleaved 3 questions
1§03.4 — a throw off a cliff quoted in kilometres per hour●●●○○
Mixed practice, and the type is not announced. A stone is thrown horizontally from the edge of a low cliff, and the throwing speed has been recorded by a radar gun that reports in kilometres per hour.
Given
Cliff height above the beach: $12.0\ \mathrm{m}$
Launch speed as recorded: $54.0\ \mathrm{km/h}$, horizontal
$g = 9.80\ \mathrm{m/s^{2}}$, air resistance neglected
Find
(a) How far from the base of the cliff does the stone land, in metres?
(b) Quote the answer with the number of significant figures the data support.
Hint 1/4
Look at the units before the physics: two different unit systems are present in the same problem and only one of them can be used with the value of the free fall constant given.
Hint 2/4
Convert with a factor equal to one, $1\ \mathrm{km/h} = 1000/3600\ \mathrm{m/s}$, then apply the projectile split: $t = \sqrt{2h/g}$ and $x = v_{x0}t$.
Hint 3/4
Substitute the data from the question, $h = 12.0\ \mathrm{m}$ and $54.0\ \mathrm{km/h} = 15.0\ \mathrm{m/s}$: $t = \sqrt{24.0/9.80} = 1.565\ \mathrm{s}$, then $x = 15.0\times 1.565$.
Hint 4/4
So the stone lands $23.5\ \mathrm{m}$ from the base, to three significant figures.
Show solution
The conversion is done first and once, before any physics, so that the rest of the working is in a single consistent system and no factor of 3.6 can appear halfway through.
rounded once, at the end, to the three figures the weakest datum supports
Answer $$\boxed{x = 23.5\ \mathrm{m}}$$
Check
Independent check by orders of magnitude: 15 m/s is about 15 m per second and the fall is about 1.6 s, so the answer has to be near 24 m. Independent check on the time: $v_y^{2} = 2gh = 235.2$, so $v_y = 15.34$ m/s at impact, and $15.34/9.80 = 1.565$ s, matching by a route that never used the height equation for the time.
The impact speed here is $\sqrt{15.0^{2}+15.34^{2}} = 21.5$ m/s, and it happens to be a nearly 45 degree arrival because the horizontal speed and the acquired vertical speed came out almost equal.
2§03.4 — a ball fired straight up from a moving cart●●●●○
Mixed practice, and the type is not announced. A cart runs along a straight level track at a steady speed, and a spring launcher bolted to it fires a ball vertically upward as the cart passes a marker.
Given
Cart speed relative to the ground: $5.00\ \mathrm{m/s}$, constant
Ball speed relative to the cart at launch: $12.0\ \mathrm{m/s}$, straight up
$g = 9.80\ \mathrm{m/s^{2}}$, air resistance neglected, launcher and catcher at the same height
Find
(a) How long is the ball in the air?
(b) How far does the cart travel in that time, and where does the ball land relative to the cart?
(c) What are the speed and direction of the ball at launch as seen from the ground?
Hint 1/4
Two observers are watching the same ball; decide which observer each part of the question is asking about before writing anything down.
Hint 2/4
In the ground frame $\vec{v}_{BG} = \vec{v}_{BC}+\vec{v}_{CG}$, and the vertical column obeys $T = 2v_{y0}/g$ while the horizontal column obeys $x = v_{x0}t$ with $v_{x0}$ constant.
Hint 3/4
Substitute the data from the question, $v_{BC} = 12.0\ \mathrm{m/s}$ upward and $v_{CG} = 5.00\ \mathrm{m/s}$ horizontal: the ground frame launch velocity is $(5.00,\ 12.0)$ m/s and $T = 2(12.0)/9.80$.
Hint 4/4
So the flight lasts $2.45\ \mathrm{s}$, the cart moves $12.2\ \mathrm{m}$, the ball lands back in the launcher, and the ground frame launch speed is $13.0\ \mathrm{m/s}$ at $67.4^{\circ}$ above the horizontal.
Show solution
The flight time is computed in the ground frame, because that is where the free fall constant is quoted, and only the horizontal comparison is then made between the two frames.
nothing changed the ball's horizontal velocity either, so the two horizontal displacements are identical
$$x_{\text{ball}}-x_{\text{cart}} = 0$$
the ball lands back in the launcher, which is what an observer riding the cart would have predicted by seeing it go straight up and come straight down
Answer $$\boxed{T = 2.45\ \mathrm{s},\quad x_{\text{cart}} = 12.2\ \mathrm{m},\quad \text{ball lands in the launcher},\quad \vec{v}_{BG} = 13.0\ \mathrm{m/s}\ \text{at}\ 67.4^{\circ}}$$
Check
Independent check from the cart's own frame: an observer riding the cart sees a purely vertical throw at 12.0 m/s, which returns to the launcher after $2(12.0)/9.80 = 2.45$ s. The two frames agree on the time and on the catch, and disagree only about the shape of the path, which is a straight line up and down for one and a parabola for the other.
Two observers, two different trajectories, one physical event. Neither is more correct than the other, and the shared answers are exactly the ones that do not involve the horizontal direction.
3§03.5 — rejecting a misremembered range formula●●●○○
Mixed practice, and the type is not announced. In an exam a student writes down the range formula from memory with the free fall constant squared in the denominator, and has to decide in ten seconds whether that can possibly be right.
(a) Use a dimensional check to decide whether the candidate formula can be correct.
(b) Write the correct formula and evaluate the range for the numbers given.
Hint 1/4
Do not start by comparing the formula with anything you remember; ask instead what kind of quantity each side of it is.
Hint 2/4
A correct equation has the same dimensions on both sides. Here the left side is a length, so evaluate the dimensions of the right side and compare, remembering that a sine is dimensionless.
Hint 3/4
Substitute the dimensions of the given quantities, $[v_0] = \mathrm{L\,T^{-1}}$ and $[g] = \mathrm{L\,T^{-2}}$: the candidate gives $\mathrm{L^{2}T^{-2}}/\mathrm{L^{2}T^{-4}} = \mathrm{T^{2}}$, which is a time squared and not a length.
Hint 4/4
So the candidate is rejected, the correct formula has a single $g$, and with $v_0 = 25.0\ \mathrm{m/s}$ and $\theta_0 = 40.0^{\circ}$ it gives $62.8\ \mathrm{m}$.
Show solution
The dimensional test is applied before any arithmetic, because it costs one line and rules the formula out completely, whereas plugging numbers in would produce a plausible looking value with no way to tell it was wrong.
the doubled angle is 80.0 degrees, and the formula is valid here because the problem states level ground
$$= \frac{615.5}{9.80} = 62.8\ \mathrm{m}$$
three significant figures, matching the data
Answer $$\boxed{\text{candidate rejected};\qquad R = 62.8\ \mathrm{m}}$$
Check
Independent check on the number by the two column route, which does not use the range formula at all: $v_{x0} = 25.0\cos 40.0^{\circ} = 19.15$ m/s, $v_{y0} = 25.0\sin 40.0^{\circ} = 16.07$ m/s, $T = 2(16.07)/9.80 = 3.280$ s, and $R = (19.15)(3.280) = 62.8$ m.
Note what the dimensional check could not do: it would have passed a formula with the sine replaced by a cosine, or with the whole thing multiplied by seven. It rejects, it never confirms, and that is still worth ten seconds in an exam.
Mistake ledger (18 entries)
⚠ Adding the magnitudes and calling the result a resultant
the two numbers are sitting there and the plus sign works on every other quantity in the course, so the hand does it before the head objects
the word change suggests before and after in that reading order, and the subtraction gets written in the order the words appeared rather than final minus initial
wrong$$\Delta\vec{v} = \vec{v}_i - \vec{v}_f$$
right$$\Delta\vec{v} = \vec{v}_f - \vec{v}_i$$
⚠ Turning an arrow while sliding it
the tail to tip construction involves moving the second arrow, and it is easy to also swing it round so that it meets the first one neatly
wrong$$\text{slide and rotate }\vec{B}\ \Rightarrow\ \text{a different vector}$$
right$$\text{slide }\vec{B}\ \text{only: length and direction both unchanged}$$
⚠ Using the angle to the nearest axis instead of the angle from the positive x axis
problems often quote an angle above the horizontal or west of north, and that quoted angle gets fed straight into the cosine without being converted
⚠ Letting the acceleration into the horizontal equation
the position equation was memorised as one formula with three terms, and it gets written out in full on both axes without asking what the acceleration is on each
wrong$$x = v_{x0}t+\tfrac{1}{2}gt^{2}$$
right$$x = v_{x0}t \qquad (a_x = 0)$$
⚠ Using the launch speed as the initial vertical velocity
the number given in the problem is a single speed, and it gets dropped into whichever column is being written first
⚠ Taking the time of flight as twice the time to the apex when the ground is lower
the symmetry of the parabola is genuinely true for a level flight and it is a very useful shortcut there, so it gets carried into the cliff and roof problems where it silently fails
the two orders read almost identically out loud, and the wrong one still produces a plausible looking triangle
wrong$$\vec{v}_{BG} = \vec{v}_{BW}+\vec{v}_{GW}$$
right$$\vec{v}_{BG} = \vec{v}_{BW}+\vec{v}_{WG}$$
⚠ Using the ground speed to find the crossing time
the ground speed is the biggest number in the problem and it is the speed the boat really travels at, so it feels like the one that should divide the width
frames moving at constant velocity relative to one another, at everyday speeds; the inner subscripts must match
River crossing results
$$t_{\text{across}} = \frac{w}{(v_{BG})_{\perp}},\qquad \sin\theta = \frac{v_{WG}}{v_{BW}}\ \text{to land straight across}$$
the second result needs $v_{BW} > v_{WG}$, otherwise the crossing straight across is impossible
Check yourself
Close the page and write down, from memory: what makes a quantity a vector; the two extremes between which the magnitude of a sum has to lie; the two formulas that resolve a vector and the two that rebuild it, together with what the arctangent cannot tell you; the difference between average velocity and average speed and which of them is always larger; the two projectile columns with the acceleration in each; what is true at the highest point and what is not; the three level ground results and the one condition they all carry; and the subscript rule for relative velocity. Then reopen and compare. The gaps are your reread list, and none of this is scored.
Draw two vectors tail to tip, say what the largest and smallest possible magnitudes of their sum are, and find the angle between two legs from the length of their resultant?
c-vectors-and-scalars
Resolve a vector at 145 or 210 degrees with the right signs, add three such vectors by columns, and repair an arctangent that has landed in the wrong quadrant?
c-vector-components
Differentiate a position vector component by component, and say without hesitating which of average speed and average velocity magnitude is the larger and why?
c-vector-kinematics
Write the two projectile columns from memory with the correct acceleration in each, and explain why a horizontally launched ball and a dropped ball land together?
c-projectile-motion
Find height, flight time, range and impact velocity for a launch at an angle, and recognise in one glance when the range formula must be abandoned for a quadratic?
c-projectile-problems
Build a relative velocity by matching inner subscripts, and get both the drift for a straight across crossing and the heading needed to land directly opposite?
c-relative-velocity
Glossary (22 terms)
vectorvektör
A quantity that needs a direction as well as a size to be specified, and that combines with others by the tail to tip construction rather than by ordinary addition.
scalarskaler
A quantity fully specified by a number and a unit, with no direction attached, such as mass, time, distance or speed.
magnitudebüyüklük
The length of a vector, always positive or zero, carrying the unit of the quantity itself.
componentbileşen
The signed projection of a vector onto one of the chosen axes; two of them carry exactly as much information as the arrow.
unit vectorbirim vektör
A vector of length one and no unit, used only to name a direction, written with a hat as in i hat and j hat.
resultantbileşke
The single vector equivalent to a set of vectors added together, drawn from the first tail to the last head.
position vectorkonum vektörü
The arrow from the chosen origin to where an object is at a given instant.
displacementyer değiştirme
The change in the position vector between two instants, which depends only on the two endpoints and not on the path taken.
alınan yol
The length of the path actually followed, a scalar that is never less than the magnitude of the displacement.
average velocityortalama hız
The displacement divided by the elapsed time, a vector pointing along the straight line from start to finish.
instantaneous velocityanlık hız
The derivative of the position vector with respect to time, a vector tangent to the path at that instant.
average speedortalama sürat
The distance travelled divided by the elapsed time, a scalar that is at least as large as the magnitude of the average velocity.
average accelerationortalama ivme
The change in the velocity vector divided by the elapsed time; it can be non zero even when the speed never changes.
projectileatılan cisim
An object moving freely near the ground after release, with air resistance neglected, so that its only acceleration is the constant downward one.
trajectoryyörünge
The path traced out by a moving object, which for a projectile in this model is a parabola.
rangemenzil
The horizontal distance from launch to landing; the compact formula for it holds only when the two heights are equal.
time of flightuçuş süresi
The interval between launch and landing, fixed entirely by the vertical column of the motion.
launch angleatış açısı
The angle between the initial velocity and the horizontal, which decides how the launch speed is divided between the two columns.
free fall accelerationyer çekimi ivmesi
The constant downward acceleration of magnitude 9.80 metres per second squared shared by all freely moving objects near the surface of the Earth.
relative velocitybağıl hız
The velocity of one object as measured by an observer who is themselves moving, written with two ordered subscripts.
frame of referencereferans sistemi
The observer, together with the axes and clock they use; two frames moving relative to one another disagree about velocities but agree about times and about what happened.
paralelkenar yöntemi
A construction for adding two vectors by drawing them from a common tail and taking the diagonal of the parallelogram they span.
What comes next
§04 · Dynamics: Newton's Laws of Motion
Everything in this section describes motion without ever asking what caused it: the acceleration of a projectile was handed to us as a measured fact about objects near the ground, and every problem started after the ball had already left the hand. Next week the question changes from how things move to what makes them move that way, and the vector machinery built here is what that answer will be written in. Nothing you have just learned gets put away; components, resultants and the habit of splitting a plane problem into two axis problems are used on the very first page.
Sources
D. Giancoli, Physics for Scientists and Engineers with Modern Physics, 5th edition, the chapter carrying the same title as this week's syllabus line The required book for the course. The treatment of vectors, components, vector kinematics, projectile motion and relative velocity follows this text. No section numbers are quoted anywhere in these notes, because the week line in the syllabus gives none and inventing them would be worse than leaving them out.
PHYS 101 syllabus: the week three line and the published assessment weights The week line reads Kinematics in Two or Three Dimensions; Vectors, and the coverage table maps that line onto the concepts of this section. The weights quoted in the sixty second card are the published ones and nothing beyond them is claimed.
The International System of Units, for the metre, the second and the derived units of velocity and acceleration Used only for the unit symbols and the conversion factors between kilometres per hour and metres per second, all of which are exact by definition.