Exercise 1: Two columns and one clock

A steel ball rolls along a horizontal bench and leaves the edge at $v_{0} = 4.0$ m/s. The bench top is $1.225$ m above the floor. Take $g = 9.8$ m/s $^{2}$ and ignore the air, as this course always does.

Before writing a single number, rule the page into two columns. The horizontal column holds a motion at constant velocity, since nothing pushes the ball sideways once it has left the bench. The vertical column holds a free fall, the one you met in the previous chapter. The two columns share exactly one quantity, the time.

a) How long does the ball stay in the air?
b) How far from the foot of the bench does it land?
c) Give both components of the velocity at impact, then the speed and the angle below the horizontal.
d) A second steel ball, held at rest against the edge at the same height, is let go at the very instant the first one rolls off. Compare the instants at which the two balls hit the floor.
e) Now compare their speeds just before impact. Give the speed of the ball that was let go, and say which ball, if either, strikes the floor faster.

Show the solution

Answers

a) $t = 0.50$ s
b) $x = 2.0$ m
c) $v_{x} = 4.0$ m/s and $v_{y} = 4.9$ m/s downward, so $v = 6.3$ m/s at $50.8^{\circ}$ below the horizontal
d) They land together, since the vertical column is identical for both and the horizontal speed never enters it
e) The ball that rolled off, $6.3$ m/s against $4.9$ m/s: same vertical component, plus a horizontal one the other ball does not have

The two columns, written first, decide everything that follows. Horizontal: $v_{x} = 4.0$ m/s, constant, no acceleration. Vertical: $v_{0y} = 0$ (the ball leaves horizontally), $a_{y} = g = 9.8$ m/s $^{2}$ downward, drop $= 1.225$ m. Taking down as positive in the vertical column keeps every vertical number positive here, which removes half the sign errors.

a)

The height lives in the vertical column, so the time comes from the vertical column alone: $1.225 = \frac{1}{2}(9.8)t^{2}$ , hence $t^{2} = \frac{2(1.225)}{9.8} = 0.25$ and $t = 0.50$ s. Look at what did NOT appear in that line: the $4.0$ m/s. A ball leaving the same bench at $40$ m/s would still be in the air for $0.50$ s.

b)

Now, and only now, the time crosses into the horizontal column: $x = v_{x}t = (4.0)(0.50) = 2.0$ m.

c)

The horizontal component never changes: $v_{x} = 4.0$ m/s. The vertical one is built by gravity: $v_{y} = gt = (9.8)(0.50) = 4.9$ m/s downward. The speed is the length of the vector, $v = \sqrt{4.0^{2} + 4.9^{2}} = \sqrt{40.01} = 6.3$ m/s, and its direction is given by $\tan\theta = \frac{4.9}{4.0} = 1.225$ , so $\theta = 50.8^{\circ}$ below the horizontal.

Numerical check without redoing part a): the vertical column also gives $v_{y}^{2} = 2g\,\Delta y = 2(9.8)(1.225) = 24.01$ , so $v_{y} = 4.9$ m/s. Two independent routes to the same number is the cheapest verification in this chapter, and it costs ten seconds.

d)

They land together. The ball that was let go has $v_{0y} = 0$ too, the same $g$ and the same $1.225$ m, so its vertical column is word for word the one written in part a). The rolled ball simply travels $2.0$ m sideways while it falls. This is the whole chapter in one sentence: the horizontal motion does not slow the fall and the fall does not slow the horizontal motion.

e)

Same landing time, different speeds, and there is no contradiction between the two answers.

Both vertical components reach $4.9$ m/s downward, since part d) showed that the two vertical columns are identical.

The ball that was let go has nothing else, so it strikes the floor at $4.9$ m/s.

The rolled ball still carries its $4.0$ m/s sideways, so it strikes at the $6.3$ m/s of part c).

A speed is the length of the whole vector, $\sqrt{v_{x}^{2} + v_{y}^{2}}$ , and only one of the two balls has a horizontal component.

The energy route reaches the same verdict in one line, $v^{2} = v_{0}^{2} + 2gh$ : $0 + 24.01$ against $16 + 24.01$ .

The trap and its price. The line that appears on weak copies is $1.225 = (4.0)t$ , or worse, $\sqrt{2.0^{2} + 1.225^{2}}$ used as a distance in a free fall equation. Both mix a horizontal number into a vertical equation, and both destroy every answer that follows: on a five part question that is the whole question, not one point. The rule that prevents it: a quantity written in the vertical column must have been measured vertically, and the only quantity allowed to cross the line between the columns is $t$ .

d) Both balls flashed every

0.1

s: at every flash they are at the same height, joined by a dashed line, so they land together at

t = 0.50

s. Only the rolled ball moves sideways,

0.4

m per flash.

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Exercise 2: Launched at an angle, and the apex that is not a stop

A ball is launched from ground level at $v_{0} = 25$ m/s, at an angle $\theta$ above the horizontal such that $\sin\theta = 0.60$ and $\cos\theta = 0.80$ , that is $\theta = 36.9^{\circ}$ . It lands on the same horizontal ground. Take $g = 9.8$ m/s $^{2}$ .

Everything changes from the previous exercise except the method: the vertical column now starts with a non zero speed. Resolve the launch velocity into components before anything else, then keep the two columns apart.

a) Find the two components of the launch velocity.
b) How long does the ball take to reach its highest point?
c) What is the maximum height reached?
d) Give the velocity vector at the highest point, magnitude and direction.
e) Find the total time of flight and the range.

Show the solution

Answers

a) $v_{0x} = 20$ m/s and $v_{0y} = 15$ m/s
b) $t_{\text{up}} = 1.53$ s
c) $H = 11.5$ m
d) $20$ m/s, horizontal, pointing forward: only the vertical component is zero
e) $T = 3.06$ s and $R = 61.2$ m

a)

The components come from the right triangle of Chapter 1: $v_{0x} = v_{0}\cos\theta = (25)(0.80) = 20$ m/s and $v_{0y} = v_{0}\sin\theta = (25)(0.60) = 15$ m/s. Check the triangle closes: $\sqrt{20^{2} + 15^{2}} = \sqrt{625} = 25$ m/s.

b)

Rising, slowing, stopping in the vertical sense: that is the free fall of the previous chapter, with $v_{y} = v_{0y} - gt$ if up is positive. At the top $v_{y} = 0$ , so $t_{\text{up}} = \frac{v_{0y}}{g} = \frac{15}{9.8} = 1.53$ s.

c)

$H = \frac{v_{0y}^{2}}{2g} = \frac{15^{2}}{2(9.8)} = \frac{225}{19.6} = 11.5$ m. Equivalently, $H = v_{0y}t_{\text{up}} - \frac{1}{2}g\,t_{\text{up}}^{2} = (15)(1.531) - (4.9)(1.531)^{2} = 11.5$ m. Notice again that $v_{0x}$ appears nowhere: a ball launched with the same $15$ m/s upward but only $2$ m/s forward reaches exactly the same height.

d)

This is the single most expensive line of the chapter. At the top the ball is NOT at rest. Its vertical component is zero, its horizontal component is what it has always been: $v_{x} = 20$ m/s. So the velocity at the apex is $20$ m/s, horizontal, pointing forward. If it really were zero the ball would drop vertically from there, and the second half of the trajectory would be a straight line down rather than the mirror image of the first half.

e)

Same ground, so the flight is symmetric: it takes as long to come down as to go up, $T = 2t_{\text{up}} = 3.06$ s. The time then crosses into the horizontal column: $R = v_{0x}T = (20)(3.06) = 61.2$ m. The compact form, worth memorising, is $R = \frac{v_{0}^{2}\sin 2\theta}{g} = \frac{625\sin(73.8^{\circ})}{9.8} = 61.2$ m, but it is valid ONLY when launch and landing are at the same height, which is the case here and will not be the case in exercise 6.

Verification in five seconds: the ball is in the air for about $3$ s and moves forward at $20$ m/s, so the range must be around $60$ m. A student who writes $R = 306$ m has multiplied by the wrong time, and a student who writes $R = 11.5$ m has copied the height into the range.

The trap and its price. Writing $v = 0$ at the apex costs the mark for part d) and usually wrecks part e) as well, because the same student then computes the second half of the flight as a drop from rest at $x = 30.6$ m, which shortens the range by half. Say it out loud once and it stays: at the top, the ball still has all of its horizontal speed, and gravity only ever eats the vertical one.

The velocity at launch, at the top and at landing, with its components:

v_{x} = 20

m/s never changes,

v_{y}

goes from

+15

to

0

to

-15

m/s, and the acceleration

g

is the same brown arrow at every point, apex included.

Exercise 3: A strobe photograph, read column by column

A ball is launched over a bench and photographed by a strobe lamp flashing every $0.100$ s. The table gives the position of the ball in each flash, measured from the launch point, and the figure plots the same six positions.

No equation of the chapter is given to you here. Everything is read off the two columns of numbers, which is exactly what a laboratory measurement looks like.

$t$ (s)	$x$ (m)	$y$ (m)
0.000	0.000	0.000
0.100	0.600	0.441
0.200	1.200	0.784
0.300	1.800	1.029
0.400	2.400	1.176
0.500	3.000	1.225

a) Show that the horizontal motion is uniform and give $v_{x}$ .
b) Using the successive differences of the $y$ column, find the vertical acceleration.
c) Find the vertical component of the launch velocity, then the launch speed and angle.
d) Predict the time and the horizontal distance at which the ball returns to the height it started from.

Show the solution

Answers

a) The gaps in $x$ are all $0.600$ m, so $v_{x} = 6.00$ m/s
b) The second differences are all $-0.098$ m, so $g = 9.8$ m/s $^{2}$ downward
c) $v_{0y} = 4.90$ m/s, so $v_{0} = 7.75$ m/s at $39.2^{\circ}$ above the horizontal
d) $t = 1.00$ s and $x = 6.00$ m

a)

Take the gaps in the $x$ column: $0.600$ , $0.600$ , $0.600$ , $0.600$ , $0.600$ m. Equal distances in equal times is the definition of uniform motion, so $v_{x} = \frac{0.600}{0.100} = 6.00$ m/s, and that number will not change for the whole flight.

b)

The $y$ gaps are NOT equal: $0.441$ , $0.343$ , $0.245$ , $0.147$ , $0.049$ m. They shrink by the same amount each time, $-0.098$ m, and that constant second difference is the signature of a constant acceleration. The relation is $\Delta^{2}y = a\,(\Delta t)^{2}$ , so $a = \frac{-0.098}{(0.100)^{2}} = -9.8$ m/s $^{2}$ , that is $9.8$ m/s $^{2}$ downward. This is how a laboratory measures $g$ without ever differentiating anything.

c)

Over the first interval the AVERAGE vertical velocity is $\frac{0.441}{0.100} = 4.41$ m/s. For a constant acceleration the average velocity over an interval equals the velocity at its midpoint, here $t = 0.050$ s, so $v_{0y} = 4.41 + (9.8)(0.050) = 4.90$ m/s. The direct route gives the same thing: $0.441 = v_{0y}(0.100) - (4.9)(0.100)^{2}$ , so $v_{0y} = \frac{0.441 + 0.049}{0.100} = 4.90$ m/s.

Then the launch vector is rebuilt from its two components: $v_{0} = \sqrt{6.00^{2} + 4.90^{2}} = \sqrt{60.01} = 7.75$ m/s, at $\theta = \arctan\frac{4.90}{6.00} = 39.2^{\circ}$ above the horizontal.

Reading the table confirms it: the ball reaches its highest point at $t = \frac{v_{0y}}{g} = \frac{4.90}{9.8} = 0.500$ s, and indeed the last row, $y = 1.225$ m, is where the $y$ column stops climbing. The maximum height also matches, $\frac{4.90^{2}}{2(9.8)} = 1.225$ m.

d)

Back to the launch height means $y = 0$ again. By symmetry that is twice the time to the top, $t = 1.00$ s, and the time then crosses into the horizontal column: $x = (6.00)(1.00) = 6.00$ m. Note that the table alone would never have told you this: the photograph stops at the apex, and the prediction comes from the model.

The trap and its price. The figure is a picture of the PATH, $y$ against $x$ , not a graph of the motion in time, and near the top the dots bunch up in $y$ only.

Two wrong conclusions are drawn from that bunching. The first: the ball stops at the top. It does not, it still covers $0.600$ m per flash there, so it moves at $6.00$ m/s.

The second: the horizontal speed drops near the top. The $x$ gaps say the opposite, they are all equal from the first flash to the last.

What IS true is that the ball slows down, from $7.75$ m/s at launch to $6.00$ m/s at the apex, and only because its vertical component dies away while the horizontal one never changes.

Reading the path as a time graph costs part a) and part b) together, and it is the same confusion that makes a student write the slope of the trajectory where a velocity belongs.

Exercise 4: Two angles, one range

A ball is launched from the ground at $v_{0} = 20$ m/s and lands on the same ground. Take $g = 9.8$ m/s $^{2}$ .

When launch and landing are at the same height, the range can be written in one line: $R = \frac{v_{0}^{2}\sin 2\theta}{g}$ . That formula is worth having, provided you also know what it does NOT say.

a) Compute the range for $\theta = 30^{\circ}$ and for $\theta = 60^{\circ}$ , and explain the result without computing anything.
b) Compute the range for $\theta = 45^{\circ}$ and explain why no other angle beats it.
c) The two shots of part a) land in the same place. Compare their times of flight and their maximum heights.
d) A target sits $45$ m away on the same ground. Can it be hit at $20$ m/s? What is the smallest launch speed that reaches it?
e) Two identical ball machines stand side by side on a training field, each throwing at $20$ m/s. One is set at $25^{\circ}$ , the other at $65^{\circ}$ , and a single button triggers both. Show that the two balls come down on the same spot and give its distance. Which ball reaches that spot first, and how many seconds separate the two arrivals?
f) To shrink that gap, the coach moves the target cone farther away and resets both machines to the pair of complementary angles that reaches it. What is the gap when the cone stands at $40.8$ m? And at $45$ m?
g) Now the general case, in letters. On a public square, two nozzles side by side are fed by the same pump, so each fires a short pulse of water at the same speed $v_{0}$ , one at $\theta > 45^{\circ}$ and the other at $90^{\circ} - \theta$ . Both pulses are fired together and come down in the same small bowl at a distance $D$ , level with the nozzles, so the bowl splashes twice. Find the ratio of the two flight times in terms of $\theta$ , and the time $\Delta t$ between the two splashes in terms of $v_{0}$ , $D$ and $g$ only (hint: square $\sin\theta - \cos\theta$ ). Check its unit and its two limits, $D \to \frac{v_{0}^{2}}{g}$ and $D \to 0$ . Then, with $v_{0} = 9.0$ m/s and $g = 9.81$ m/s $^{2}$ as on the formula sheet, find where the bowl must stand for the two splashes to be $0.50$ s apart, the angle of the steep nozzle, and the ratio of the flight times.

Type your answers, the page tells you right or wrong 0/17

a)

Range at

30^{\circ}

m

Range at

60^{\circ}

m

b)

Range at

45^{\circ}

m

c)

T

at

30^{\circ}

s

T

at

60^{\circ}

s

H

at

30^{\circ}

m

H

at

60^{\circ}

m

d)

Can the target be hit at

20

m/s?

Smallest launch speed m/s

e)

Distance of the landing spot m

First ball to land

Gap between the arrivals s

f)

Gap at

40.8

m s

At

45

m

g)

Distance

D

of the bowl m

\theta

, steep nozzle °

\frac{T_{\text{steep}}}{T_{\text{flat}}}

Show the solution

Answers

a) $R = 35.3$ m for both, because $\sin 60^{\circ} = \sin 120^{\circ}$
b) $R = 40.8$ m, the largest possible, because $\sin 2\theta$ peaks at $2\theta = 90^{\circ}$
c) $T = 2.04$ s and $H = 5.10$ m at $30^{\circ}$ , against $T = 3.53$ s and $H = 15.3$ m at $60^{\circ}$
d) No, the best possible range at $20$ m/s is $40.8$ m. The smallest launch speed is $21$ m/s, at $45^{\circ}$
e) Same spot, $31.3$ m away; the $25^{\circ}$ ball lands first, $1.97$ s ahead of the $65^{\circ}$ one
f) At $40.8$ m both machines sit at $45^{\circ}$ and the gap is zero; at $45$ m neither ball reaches the cone
g) $\frac{T_{\text{steep}}}{T_{\text{flat}}} = \tan\theta$ and $\Delta t = \frac{2}{g}\sqrt{v_{0}^{2} - gD}$ ; for $0.50$ s, $D = 7.64$ m, $\theta = 56.1^{\circ}$ and a ratio of $1.49$

a)

At $30^{\circ}$ : $R = \frac{20^{2}\sin 60^{\circ}}{9.8} = \frac{400(0.866)}{9.8} = 35.3$ m. At $60^{\circ}$ : $R = \frac{20^{2}\sin 120^{\circ}}{9.8} = 35.3$ m. The two are equal and no calculation was needed to know it: doubling the angles gives $60^{\circ}$ and $120^{\circ}$ , which are supplementary, and supplementary angles have the same sine. Any two launch angles that ADD UP TO $90^{\circ}$ share a range, which is why $15^{\circ}$ and $75^{\circ}$ , or $40^{\circ}$ and $50^{\circ}$ , are also twins.

b)

At $45^{\circ}$ : $R = \frac{400\sin 90^{\circ}}{9.8} = \frac{400}{9.8} = 40.8$ m. No angle does better, and the reason is visible in the formula rather than hidden in a derivative: $R$ is proportional to $\sin 2\theta$ , and a sine never exceeds $1$ . It equals $1$ only when $2\theta = 90^{\circ}$ , that is $\theta = 45^{\circ}$ . This is the one optimisation in the chapter that needs no calculus at all.

c)

Same landing point, completely different flights. Times: $T = \frac{2v_{0}\sin\theta}{g}$ gives $2.04$ s at $30^{\circ}$ and $3.53$ s at $60^{\circ}$ , a ratio of $\frac{\sin 60^{\circ}}{\sin 30^{\circ}} = \sqrt{3}$ . Heights: $H = \frac{(v_{0}\sin\theta)^{2}}{2g}$ gives $5.10$ m and $15.3$ m, a ratio of exactly $3$ , since the height goes with the SQUARE of the vertical component.

The steep shot wins its extra time and loses exactly as much horizontal speed, and the two effects cancel in the product $v_{0x}T$ . That cancellation is the whole content of the twin angle result, and it is also why the two shots are useful for different purposes: the flat one arrives sooner, the steep one arrives more vertically and clears an obstacle standing in the middle of the field.

d)

The maximum range at $20$ m/s is the $40.8$ m of part b), so $45$ m is out of reach at ANY angle. To reach $45$ m the best angle is again $45^{\circ}$ , where $R = \frac{v_{0}^{2}}{g}$ , so $v_{0}^{2} = Rg = (45)(9.8) = 441$ and $v_{0} = 21$ m/s. One extra metre per second of launch speed buys four extra metres of range here, because the range grows with the SQUARE of the speed.

e)

The two settings add up to $90^{\circ}$ , so part a) applies: $\sin 50^{\circ} = \sin 130^{\circ}$ and the two balls share one range.

$R = \frac{20^{2}\sin 50^{\circ}}{9.8} = \frac{400(0.766)}{9.8} = 31.3$ m

Their clocks differ: $T = \frac{2v_{0}\sin\theta}{g}$ gives $\frac{40\sin 25^{\circ}}{9.8} = 1.725$ s and $\frac{40\sin 65^{\circ}}{9.8} = 3.699$ s.

So the flat ball is on the spot first, and the steep one lands $3.699 - 1.725 = 1.97$ s later, from much higher up.

Same speed, same button, same landing spot, and yet almost two seconds apart: the range hides the clock, which is exactly what part c) found.

f)

The gap is the difference of two flight times, $\frac{2v_{0}}{g}(\sin\theta - \cos\theta)$ for the steeper setting $\theta$ . Moving the cone away brings both settings towards $45^{\circ}$ , so the bracket, and the gap, shrink.

At $40.8$ m, the maximum range, both machines must sit at $45^{\circ}$ : the two throws are identical and the gap is zero.

At $45$ m, part d) already showed that no setting reaches the cone: neither ball gets there, and there is no gap to speak of.

The trap and its price. Two of them travel together. First, applying $R = \frac{v_{0}^{2}\sin 2\theta}{g}$ when the landing height differs from the launch height: the formula is then simply false, and exercise 6 shows by how much. Second, concluding from equal ranges that the two flights are equivalent: a question asking which shot clears a $10$ m wall midway has a unique answer, $60^{\circ}$ , and the twin angle result says nothing about it. Both mistakes cost the full mark of the question they touch, because the number produced is not approximately right, it is about something else.

g)

Same speed, same level at both ends, so both flights are symmetric and the two compact formulas are allowed.

$T_{\text{steep}} = \frac{2v_{0}\sin\theta}{g}$ and $T_{\text{flat}} = \frac{2v_{0}\sin(90^{\circ} - \theta)}{g} = \frac{2v_{0}\cos\theta}{g}$

Ratio: $\frac{T_{\text{steep}}}{T_{\text{flat}}} = \frac{\sin\theta}{\cos\theta} = \tan\theta$ , which gives back the $\sqrt{3} = \tan 60^{\circ}$ of part c).

The flat pulse splashes first, and the gap is the difference of the two times: $\Delta t = \frac{2v_{0}}{g}(\sin\theta - \cos\theta)$ , positive since $\theta > 45^{\circ}$ .

To remove $\theta$ , square the bracket and use the range equation, $\sin 2\theta = \frac{gD}{v_{0}^{2}}$ :

\displaystyle (\sin\theta - \cos\theta)^{2} = 1 - 2\sin\theta\cos\theta = 1 - \sin 2\theta = 1 - \frac{gD}{v_{0}^{2}}

so $\Delta t = \frac{2v_{0}}{g}\sqrt{1 - \frac{gD}{v_{0}^{2}}} = \frac{2}{g}\sqrt{v_{0}^{2} - gD}$ .

Unit: $\frac{\sqrt{\text{m}^{2}/\text{s}^{2}}}{\text{m/s}^{2}} = \frac{\text{m/s}}{\text{m/s}^{2}} = \text{s}$ , a time, and $gD$ is indeed a speed squared, so the subtraction under the root is legal.

Limit $D \to \frac{v_{0}^{2}}{g}$ , the maximum range: $\Delta t \to 0$ . Both nozzles end up at $45^{\circ}$ and the bowl splashes once, which is part f) found again.

Limit $D \to 0$ : $\Delta t \to \frac{2v_{0}}{g}$ , a jet straight up against a jet skimming the ground that lands at once: the whole flight of a vertical throw.

Sense of variation: a farther bowl means a shorter gap, a stronger pump a longer one. Applied to part e), $\frac{2}{9.8}\sqrt{20^{2} - (9.8)(31.27)} = 1.97$ s, the gap found there.

Numbers, the formula read backwards: $D = \frac{v_{0}^{2} - (g\Delta t/2)^{2}}{g} = \frac{81 - 6.02}{9.81} = 7.64$ m.

Angles: $\sin 2\theta = \frac{(9.81)(7.64)}{81} = 0.926$ , so $2\theta = 112.2^{\circ}$ (the obtuse solution, since $\theta > 45^{\circ}$ ) and $\theta = 56.1^{\circ}$ , the other nozzle at $33.9^{\circ}$ .

Ratio: $\tan 56.1^{\circ} = 1.49$ , that is $1.523$ s against $1.023$ s, indeed $0.50$ s apart.

A last check that costs nothing: the product of the two times is $\frac{4v_{0}^{2}\sin\theta\cos\theta}{g^{2}} = \frac{2D}{g} = 1.56$ s $^{2}$ , and $(1.523)(1.023) = 1.56$ s $^{2}$ .

Pitfall: taking the acute solution $2\theta = 67.8^{\circ}$ for the steep nozzle. It gives $33.9^{\circ}$ , which is the FLAT one, and the ratio comes out upside down, $0.67$ .

g) Both pulses fired at

t = 0

, a dot every

0.25

s of the same clock: the flat pulse (orange) is in the bowl at

1.02

s, the steep one (blue)

0.50

s later.

Exercise 5: Relative velocity, from a bush plane in a crosswind to a rainy windshield

Relative velocity answers one question: what does the SAME motion look like from another observer? The rule is a vector sum whose subscripts chain together, $\vec{v}_{A/C} = \vec{v}_{A/B} + \vec{v}_{B/C}$ , where the inner subscripts cancel like a fraction. Reversing an observer reverses the vector: $\vec{v}_{B/A} = -\vec{v}_{A/B}$ .

A bush plane flies at $50$ m/s RELATIVE TO THE AIR, which is what its airspeed indicator reads, while a steady wind blows FROM the east at $15$ m/s, that is towards the west. Then a car drives through vertical rain.

a) The pilot holds the nose pointed due north. Find the velocity of the plane over the ground: its magnitude, and its direction as an angle west of north.
b) A fishing camp lies $120$ km due north of the dock. Which heading, measured from north, makes the plane track straight along that line, and how many minutes does the trip then take?
c) Rain falls vertically at $8.0$ m/s in the frame of the ground, while a car drives at $15$ m/s. At what speed and in what direction does the rain strike the windshield?
d) At what speed should the car drive for the rain to arrive at $45^{\circ}$ from the vertical?

Type your answers, the page tells you right or wrong 0/9

a)

Ground speed m/s

Angle west of north °

b)

Heading east of north °

Ground speed m/s

Duration of the trip min

c)

Speed of the rain relative to the car m/s

Angle from the vertical °

The rain arrives from

d)

Speed of the car m/s

Show the solution

Answers

a) $52$ m/s, at $16.7^{\circ}$ west of north
b) Nose $17.5^{\circ}$ east of north; ground speed $47.7$ m/s, so the trip lasts $41.9$ min
c) $17$ m/s, at $61.9^{\circ}$ from the vertical, arriving from the front of the car
d) $8.0$ m/s, that is the same speed as the falling rain

a)

Sketch first, axes written on the sketch: $x$ east, $y$ north. Then the chain, read out loud: plane relative to ground equals plane relative to air plus air relative to ground.

\displaystyle \vec{v}_{P/g} = \vec{v}_{P/a} + \vec{v}_{a/g}

A wind FROM the east moves the air towards the west, so $\vec{v}_{a/g} = (-15\,;0)$ , while $\vec{v}_{P/a} = (0\,;50)$ . Hence $\vec{v}_{P/g} = (-15\,;50)$ m/s.

Magnitude: $\sqrt{15^{2} + 50^{2}} = \sqrt{2725} = 52$ m/s. Direction: $\arctan\frac{15}{50} = 16.7^{\circ}$ west of north.

The nose points north and the plane slides west, exactly like the boat of exercise 7 whose bow points across the river: pointing somewhere is not the same as going there.

Pitfall: a wind is named by where it comes FROM. Reading east as the direction the air moves puts the drift on the wrong side, with the right magnitude.

b)

Same chain, but the unknown is now the DIRECTION of $\vec{v}_{P/a}$ , whose length stays $50$ m/s, while $\vec{v}_{P/g}$ has a known direction, due north, and an unknown length. Two unknowns, two component equations.

East component: $v_{P/a,x} - 15 = 0$ , so the plane must fly $15$ m/s EAST relative to the air, into the wind.

Heading: $\sin\alpha = \frac{15}{50} = 0.30$ , so $\alpha = 17.5^{\circ}$ east of north.

North component: $v_{P/g} = 50\cos\alpha = \sqrt{50^{2} - 15^{2}} = \sqrt{2275} = 47.7$ m/s.

Trip: $t = \frac{120\,000}{47.7} = 2516$ s, that is $41.9$ min.

In the tip to tail drawing the HYPOTENUSE is now the $50$ m/s of the plane relative to the air, one leg the $15$ m/s of wind, the other the ground speed. In part a) the $50$ m/s was a leg and the ground speed the hypotenuse.

Swapping the two triangles gives $52$ m/s here, where the answer must be SMALLER than $50$ m/s, since part of the airspeed is spent cancelling the wind.

c)

Here the two velocities are not parallel, so the subtraction must be done component by component. Take east as $x$ and up as $y$ . Rain relative to ground: $(0\,;-8.0)$ . Car relative to ground: $(15\,;0)$ . Then $\vec{v}_{\text{rain}/\text{car}} = \vec{v}_{\text{rain}/\text{ground}} - \vec{v}_{\text{car}/\text{ground}} = (0 - 15\,;-8.0 - 0) = (-15\,;-8.0)$ .

Its length is $\sqrt{15^{2} + 8.0^{2}} = \sqrt{289} = 17$ m/s, and its direction is $\arctan\frac{15}{8.0} = 61.9^{\circ}$ away from the vertical, tilted towards the rear of the car. In plain terms, the rain comes at the windshield from ahead and above, which is why a stationary car gets its roof wet and a moving one gets its windshield wet. Nothing has changed about the rain: only the observer changed.

d)

The angle from the vertical satisfies $\tan\alpha = \frac{v_{\text{car}}}{v_{\text{rain}}}$ . Asking for $45^{\circ}$ is asking for $\tan\alpha = 1$ , that is $v_{\text{car}} = v_{\text{rain}} = 8.0$ m/s. At $8.0$ m/s the rain arrives exactly on the diagonal, at $\sqrt{128} = 11.3$ m/s.

The trap and its price. The subscripts are not decoration. Writing $\vec{v}_{\text{rain}/\text{car}} = \vec{v}_{\text{rain}/\text{ground}} + \vec{v}_{\text{car}/\text{ground}}$ gives $17$ m/s as well, the same length, but tilted the WRONG way, towards the front of the car, and the angle question is then lost even though the speed is right. Read the chain out loud before computing: rain relative to car equals rain relative to ground plus ground relative to car, and ground relative to car is MINUS car relative to ground. That single habit is worth two or three marks per exam paper, and it is the same habit that will decide the heading of the boat in exercise 7.

Exercise 6: Thrown from a cliff, where the symmetry is gone

From the edge of a cliff $20$ m above a beach, a ball is thrown at $v_{0} = 15$ m/s at $30^{\circ}$ above the horizontal. It lands on the sand. Take $g = 9.8$ m/s $^{2}$ .

Every shortcut of exercise 4 was built on one assumption, that the ball lands at the height it was launched from. That assumption is now false, so the shortcuts go back in the drawer and the vertical column is solved as a quadratic equation.

a) How high above the beach does the ball rise?
b) How long is it in the air?
c) How far from the foot of the cliff does it land?
d) Find the speed and the direction of the velocity at impact.
e) A shot putter in training releases the shot $h = 2.1$ m above the ground, with a speed $v_{0}$ at an angle $\theta$ above the horizontal. Before any number, find the flight time $T$ as a formula built from $v_{0}$ , $\theta$ , $h$ and $g$ . Test the formula on two situations already met in this set: $h = 0$ (exercise 4) and $\theta = 0$ (exercise 1).
f) The training video gives $v_{0} = 13.0$ m/s and $\theta = 38^{\circ}$ . Compute $T$ , then the horizontal distance from the release point to the landing point. By what percentage does that distance exceed $\frac{v_{0}^{2}\sin 2\theta}{g}$ , the level ground value for the same speed and angle?

Show the solution

Answers

a) $2.87$ m above the cliff top, that is $22.9$ m above the beach
b) $t = 2.93$ s
c) $x = 38.0$ m
d) $v = 24.8$ m/s, at $58.5^{\circ}$ below the horizontal
e) $T = \frac{v_{0}\sin\theta + \sqrt{v_{0}^{2}\sin^{2}\theta + 2gh}}{g}$ , which gives $\frac{2v_{0}\sin\theta}{g}$ when $h = 0$ and $\sqrt{\frac{2h}{g}}$ when $\theta = 0$
f) $T = 1.86$ s and $x = 19.1$ m, about $14$ percent beyond the $16.7$ m of the level ground formula

Components first: $v_{0x} = (15)\cos 30^{\circ} = 12.99$ m/s and $v_{0y} = (15)\sin 30^{\circ} = 7.50$ m/s. Take up as positive and put the origin at the throwing point, so the beach is at $y = -20$ m. Writing that minus sign down now is what makes the rest work.

a)

The rise above the launch point uses the vertical column only: $H = \frac{v_{0y}^{2}}{2g} = \frac{7.50^{2}}{19.6} = 2.87$ m, so the highest point is $22.9$ m above the sand. Note how small this is compared with the cliff: most of the flight is a fall, not a rise.

b)

The landing condition is $y = -20$ m: $-20 = 7.50t - 4.9t^{2}$ , that is $4.9t^{2} - 7.50t - 20 = 0$ . The quadratic formula gives $t = \frac{7.50 \pm \sqrt{56.25 + 392}}{9.8} = \frac{7.50 \pm 21.17}{9.8}$ , so $t = 2.93$ s or $t = -1.40$ s. The negative root is rejected: it describes where the ball WOULD have been before the throw if the same parabola had been extended backwards, which is a mathematical fact about the equation and not an event on the beach.

c)

Now the time crosses into the horizontal column: $x = v_{0x}t = (12.99)(2.93) = 38.0$ m.

d)

The horizontal component is still $12.99$ m/s. The vertical one is $v_{y} = v_{0y} - gt = 7.50 - (9.8)(2.93) = -21.17$ m/s, that is $21.17$ m/s downward. Speed: $v = \sqrt{12.99^{2} + 21.17^{2}} = \sqrt{617} = 24.8$ m/s, at $\arctan\frac{21.17}{12.99} = 58.5^{\circ}$ below the horizontal.

Independent check of that speed, without the time: $v^{2} = v_{0}^{2} + 2g h = 15^{2} + 2(9.8)(20) = 225 + 392 = 617$ , so $v = 24.8$ m/s. The check works because the height enters the vertical column the same way whatever the path taken, and it catches a wrong root instantly. It also shows that the impact speed does not depend on the launch ANGLE, only on the launch speed and the drop, while the range and the flight time depend on both.

e)

Origin at the release point, up positive, so the ground sits at $y = -h$ . The vertical column alone decides the time:

$-h = v_{0}\sin\theta\,T - \frac{1}{2}gT^{2}$ , that is $\frac{1}{2}gT^{2} - v_{0}\sin\theta\,T - h = 0$

The quadratic formula from the formula sheet, keeping the positive root since the other one is negative:

\displaystyle T = \frac{v_{0}\sin\theta + \sqrt{v_{0}^{2}\sin^{2}\theta + 2gh}}{g}

Unit: under the root, $(\text{m/s})^{2}$ and $(\text{m/s}^{2})(\text{m})$ are both $\text{m}^{2}/\text{s}^{2}$ , so the numerator is a speed and $T$ comes out in $\frac{\text{m/s}}{\text{m/s}^{2}} = \text{s}$ .

Case $h = 0$ : the root becomes $v_{0}\sin\theta$ and $T = \frac{2v_{0}\sin\theta}{g}$ , the symmetric flight of exercise 4.

Case $\theta = 0$ : $T = \frac{\sqrt{2gh}}{g} = \sqrt{\frac{2h}{g}}$ , the ball rolling off the bench in exercise 1.

Sense of variation: $h$ appears only under the root, with a plus sign, so a higher release always lengthens the flight.

f)

Components: $v_{0}\sin 38^{\circ} = 8.00$ m/s and $v_{0}\cos 38^{\circ} = 10.24$ m/s.

$T = \frac{8.00 + \sqrt{8.00^{2} + 2(9.8)(2.1)}}{9.8} = \frac{8.00 + 10.26}{9.8} = 1.86$ s

Distance: $x = (10.24)(1.863) = 19.1$ m.

Level ground value: $\frac{13.0^{2}\sin 76^{\circ}}{9.8} = \frac{169(0.970)}{9.8} = 16.7$ m.

Excess: $\frac{19.1 - 16.7}{16.7}$ , about $14$ percent. The $2.1$ m of height is worth more than $2$ m of distance, which is why a putter is coached to release high.

Pitfall: plugging these numbers into $\frac{v_{0}^{2}\sin 2\theta}{g}$ and calling it the distance. That formula assumes a landing at release height, and here it is short by $2.4$ m, far more than any rounding.

The trap and its price. The line that ruins this exercise is $T = \frac{2v_{0y}}{g} = 1.53$ s, the symmetric flight time of exercise 4, applied out of its domain. It is wrong by a factor of nearly two, it drags the range down from $38.0$ m to $19.9$ m, and it costs parts b), c) and d) together, six marks out of ten, from a single reflex. The safeguard is a question asked before choosing any formula: does the ball land at the height it left? If not, no symmetry, no $2v_{0y}/g$ , no $\frac{v_{0}^{2}\sin 2\theta}{g}$ , and the quadratic is solved in full.

Exercise 7: Crossing a river, and the heading that cancels the current

A river $80$ m wide flows at $3.0$ m/s. A boat moves at $4.0$ m/s RELATIVE TO THE WATER, which is the only speed an engine can promise. The chain of subscripts is $\vec{v}_{\text{boat}/\text{ground}} = \vec{v}_{\text{boat}/\text{water}} + \vec{v}_{\text{water}/\text{ground}}$ .

Two crossings, two different questions. In the first the pilot points the bow straight at the far bank and accepts being carried; in the second he insists on landing at the point directly opposite.

a) The bow points straight across. Find the crossing time, the drift downstream, the speed over the ground and the length of the path actually followed.
b) The pilot now wants to land directly opposite the starting point. What heading must he take, and how long does the crossing last?
c) Why is the crossing of part a) faster than the crossing of part b), even though the boat engine delivers the same $4.0$ m/s in both?
d) The current rises to $5.0$ m/s. Show that landing directly opposite has become impossible, and find the heading that makes the drift as small as it can be.

Show the solution

Answers

a) $t = 20$ s, drift $= 60$ m, ground speed $= 5.0$ m/s, path $= 100$ m
b) Bow at $48.6^{\circ}$ upstream from the across direction, and $t = 30.2$ s
c) Because in part a) the whole $4.0$ m/s is spent crossing, while in part b) only $2.65$ m/s of it is
d) Impossible since $5.0 > 4.0$ . The least drift is $60$ m, reached with the bow at $53.1^{\circ}$ upstream, crossing in $33.3$ s

a)

Left triangle of the figure. The bow points across, so the boat velocity relative to the water is $(0\,;4.0)$ with the across direction as $y$ , and the water carries everything downstream at $(3.0\,;0)$ . The crossing is decided by the ACROSS component alone, which is $4.0$ m/s, untouched by the current: $t = \frac{80}{4.0} = 20$ s. During those $20$ s the river has moved the boat $(3.0)(20) = 60$ m downstream. Ground speed: $\sqrt{4.0^{2} + 3.0^{2}} = 5.0$ m/s, and the path actually traced on the water is $(5.0)(20) = 100$ m, the hypotenuse of the $60$ by $80$ triangle.

b)

Right triangle of the figure. Landing directly opposite means the downstream component of the GROUND velocity must be zero, so the upstream part of the boat velocity has to eat the current exactly: $(4.0)\sin\theta = 3.0$ , hence $\sin\theta = 0.75$ and $\theta = 48.6^{\circ}$ upstream from the across direction. What is left for crossing is $(4.0)\cos\theta = \sqrt{4.0^{2} - 3.0^{2}} = 2.65$ m/s, so $t = \frac{80}{2.65} = 30.2$ s.

c)

The engine gives $4.0$ m/s and no more, and that budget is spent in two ways. Pointing straight across spends all of it on crossing and nothing on fighting the current, which is why it is the fastest crossing possible, $20$ s, and why it is also the one that drifts most. Aiming upstream spends $3.0$ m/s of the budget on cancelling the current and leaves only $2.65$ m/s for the crossing itself, which takes half again as long. There is no heading that is best at both, and an examiner who asks for the QUICKEST crossing is asking for the bow straight across, not for the shortest path.

d)

With a current of $5.0$ m/s, cancelling it would require $(4.0)\sin\theta = 5.0$ , that is $\sin\theta = 1.25$ , which no angle satisfies. The boat is simply slower than the river and will be carried downstream whatever it does.

The least drift is found with a compass and no calculus. Draw the current vector $(5.0\,;0)$ , then all the possible boat velocities: their tips fill a circle of radius $4.0$ centred on the tip of the current vector. The ground velocity runs from the origin to a point of that circle, and the drift angle away from the across direction is smallest when that line is TANGENT to the circle. Tangent means the radius is perpendicular to it, so the right triangle has hypotenuse $5.0$ and one leg $4.0$ : the tangent line makes an angle with the downstream direction whose sine is $\frac{4.0}{5.0}$ , that is $53.1^{\circ}$ , hence $36.9^{\circ}$ away from the across direction.

In numbers: the bow at $53.1^{\circ}$ upstream gives an across component of $(4.0)(0.60) = 2.4$ m/s and a downstream component of $5.0 - (4.0)(0.80) = 1.8$ m/s. The crossing takes $\frac{80}{2.4} = 33.3$ s and the drift is $(1.8)(33.3) = 60$ m, which is indeed $80\tan 36.9^{\circ}$ .

The trap and its price. The most frequent loss here is answering part a) with $\frac{80}{5.0} = 16$ s, using the ground speed as if it crossed the river. Only the across component crosses the river; the other component runs parallel to the banks for ever and brings the boat no closer to the far side. The second loss is adding the current to the boat speed as numbers, $4.0 + 3.0 = 7.0$ m/s: velocities add as VECTORS, and here the sum of two perpendicular vectors of $4.0$ and $3.0$ is $5.0$ , not $7.0$ .

Exercise 8: Five statements to correct, then a multiple choice question

Each statement below is either exactly right or plausibly wrong. Say which, and rewrite every false one so that it becomes true. A correction that only says false earns nothing: the mark is for the sentence that replaces it.

Take $g = 9.8$ m/s $^{2}$ and ignore the air throughout, except where a statement brings it in.

Part f) is different: a four option question of the kind that opens the midterm, and the figure belongs to it.

a) At the highest point of its flight, the velocity of a projectile is zero.
b) A garden hose nozzle is advertised as delivering its jet at $12$ m/s. To test the claim, the nozzle is held level $0.90$ m above a lawn, and the jet reaches the grass $4.7$ m away, which gives an exit speed of $11.0$ m/s. Air drag can only have carried the water further than it would have gone in vacuum, so the true exit speed is at most $11.0$ m/s and the advertisement overstates it.
c) Two projectiles launched at the same speed, one at $30^{\circ}$ and one at $60^{\circ}$ , land at the same distance, so they also spend the same time in the air.
d) A boat whose bow points straight across a river crosses it in the same time as it would in still water.
e) Doubling the launch speed doubles the range.
f) A passenger stands in a tram that runs along a straight, level track at a constant $5.0$ m/s. She lets her keys drop from her hand, $1.10$ m above the floor of the tram. Which of the four paths A, B, C, D of the figure does a person standing still on the platform see the keys follow? Then find how far the keys travel along the platform before they hit the floor, taking $g = 9.81$ m/s $^{2}$ as on the formula sheet.

Show the solution

Answers

a) False: only the vertical component is zero, the horizontal one is unchanged
b) False: the $11.0$ m/s is right without air, but drag SHORTENS the jet, so the test gives a LOWER bound and cannot refute the $12$ m/s claim
c) False: the flight times are in the ratio $\sqrt{3}$ , that is $2.04$ s against $3.53$ s at $20$ m/s
d) True
e) False: the range is multiplied by $4$ , since it goes with $v_{0}^{2}$
f) Path B, a half parabola curving forward, $2.37$ m long along the platform; path A is what the passenger sees

a)

FALSE. At the top the VERTICAL component of the velocity is zero, which is exactly what defines the top; the horizontal component has never changed and never will. Correct version: at the highest point the velocity is horizontal, of magnitude $v_{0}\cos\theta$ . For the ball of exercise 2 that is $20$ m/s, not $0$ . A zero velocity would mean the object hangs still in the air, and the rest of its path would be a vertical drop.

b)

FALSE, and the false part is the direction given to the air's effect.

The measurement itself is sound: the vertical column gives the time, $t = \sqrt{\frac{2(0.90)}{9.8}} = 0.429$ s, and the horizontal column gives the speed, $v_{0} = \frac{4.7}{0.429} = 11.0$ m/s.

But drag slows the water horizontally during its flight, so the jet lands SHORT of where a vacuum jet with the same exit speed would land.

A shorter $d$ in $v_{0} = d\sqrt{\frac{g}{2h}}$ gives a smaller $v_{0}$ : the method UNDERESTIMATES the exit speed.

Correct version: without air the water leaves at $11.0$ m/s, and with air it left at $11.0$ m/s or more, so this test gives a lower bound and cannot show that the $12$ m/s is exaggerated.

Before concluding from any such measurement, ask in which direction every neglected effect pushes the answer.

c)

FALSE, and the first half of the sentence is the reason people believe the second. The ranges are equal because $\sin 2\theta$ takes the same value at $30^{\circ}$ and at $60^{\circ}$ . The times are not, because $T = \frac{2v_{0}\sin\theta}{g}$ depends on $\sin\theta$ and not on $\sin 2\theta$ . At $20$ m/s the flat shot lasts $2.04$ s and the steep one $3.53$ s, a ratio of $\sqrt{3}$ . Correct version: equal ranges, but the steeper shot stays in the air $\sqrt{3}$ times longer and climbs three times higher.

d)

TRUE, and it surprises most students. The time to cross depends only on the component of the velocity PERPENDICULAR to the banks, and the current is parallel to them, so the current changes nothing about that component. With a $4.0$ m/s boat on an $80$ m river the crossing lasts $20$ s whether the water is still or flowing at $3.0$ m/s. What the current does change is where the boat lands, $60$ m downstream, and how long the path traced on the water is, $100$ m instead of $80$ m.

e)

FALSE. On level ground $R = \frac{v_{0}^{2}\sin 2\theta}{g}$ , and the launch speed appears SQUARED, so doubling it multiplies the range by $4$ . At $45^{\circ}$ , going from $20$ m/s to $40$ m/s takes the range from $40.8$ m to $163$ m. Correct version: doubling the launch speed multiplies the range by four, and multiplies the maximum height by four as well, while it only doubles the time of flight.

What these five have in common, and what to take into the exam: three of them treat a quantity of one column as if it belonged to the other, or a proportionality as linear when a square is involved, and the fourth misjudges the direction of a neglected effect. Before accepting any sentence about a projectile, ask which column each quantity lives in, whether the relation is linear, quadratic or trigonometric, and which way every approximation leans.

f)

Path B. At the instant she opens her hand the keys already move forward with the tram, $5.0$ m/s relative to the platform, and nothing horizontal acts on them afterwards.

So for the platform the two columns are those of exercise 1: uniform motion at $5.0$ m/s horizontally, free fall from rest vertically, and their combination is a half parabola curving forward.

$t = \sqrt{\frac{2(1.10)}{9.81}} = 0.474$ s, then $x = (5.0)(0.474) = 2.37$ m along the platform.

The tram also moves $2.37$ m in that time, so the keys land at her feet, exactly as she sees it.

A, the vertical line, is the passenger's own view: true in the frame of the tram, wrong in the frame of the platform, the same pair of descriptions as exercise 10 d).

C, curving backwards, is the belief that the keys are left behind: it would need a force pushing them backwards, and there is none, the air inside the tram moving with it.

D, the straight slanted line, keeps the uniform horizontal motion but forgets that the fall accelerates: equal steps in $x$ with growing steps in $y$ draw a curve, as the table of exercise 3 showed.

f) The keys every

0.1

s, in the tram and from the platform: on the platform the hand moves on at

5.0

m/s and the keys stay directly below it, on a forward half parabola.

Exercise 9: A grasshopper filmed at 240 frames per second

In a field study, a grasshopper $3.0$ cm long is filmed while it jumps. From the video the biologists measure two numbers only: the jump covers $0.90$ m horizontally and the animal rises $0.30$ m above the ground at the top. Take off and landing are both at ground level, and the air is ignored. Take $g = 9.8$ m/s $^{2}$ .

Nothing else is measured, and nothing else is needed: the range and the maximum height together contain the whole launch vector.

a) Show that the take off angle satisfies $\tan\theta = \frac{4H}{R}$ , and compute it.
b) Find the take off speed.
c) How long is the animal in the air, and how many frames of the video show the jump?
d) The same jump performed on the Moon, where gravity is about six times weaker, would cover what distance?
e) The same grasshopper, with the same take off speed and the same angle ABOVE THE HORIZONTAL, now jumps downhill from a point on a slope inclined at $\alpha = 30^{\circ}$ . Write the time of flight as a formula, check that it gives back the level ground flight when $\alpha = 0$ and say how it changes as the slope steepens, then compute the time and the distance measured along the slope.

Show the solution

Answers

a) $\tan\theta = \frac{4H}{R} = 1.33$ , so $\theta = 53.1^{\circ}$
b) $v_{0} = 3.03$ m/s
c) $t = 0.495$ s, that is about $119$ frames
d) About $5.4$ m, six times further
e) $t = \frac{2v_{0}(\sin\theta + \cos\theta\tan\alpha)}{g} = 0.709$ s, against $0.495$ s on level ground, and $1.49$ m along the slope

a)

Write the two level ground results side by side: $H = \frac{(v_{0}\sin\theta)^{2}}{2g}$ and $R = \frac{2v_{0}^{2}\sin\theta\cos\theta}{g}$ . Divide the first by the second and the speed disappears: $\frac{H}{R} = \frac{v_{0}^{2}\sin^{2}\theta / (2g)}{2v_{0}^{2}\sin\theta\cos\theta / g} = \frac{\sin\theta}{4\cos\theta} = \frac{\tan\theta}{4}$ , so $\tan\theta = \frac{4H}{R}$ .

With the measurements: $\tan\theta = \frac{4(0.30)}{0.90} = 1.33$ , so $\theta = 53.1^{\circ}$ . That is a steep jump, and a very common value in the literature for insects that jump to escape rather than to travel.

b)

Use the height, which involves only the vertical component: $v_{0}\sin\theta = \sqrt{2gH} = \sqrt{2(9.8)(0.30)} = 2.42$ m/s. Since $\sin 53.1^{\circ} = 0.80$ , the take off speed is $v_{0} = \frac{2.42}{0.80} = 3.03$ m/s. Check with the other measurement: $R = \frac{v_{0}^{2}\sin 2\theta}{g} = \frac{(3.03)^{2}(0.96)}{9.8} = 0.90$ m, which is the measured range. The two measurements are consistent, so the parabolic model holds for this jump.

c)

The time follows from the vertical column alone: $t = \frac{2v_{0}\sin\theta}{g} = \frac{2(2.42)}{9.8} = 0.495$ s. A shorter route avoids the speed entirely, $t = \sqrt{\frac{8H}{g}} = \sqrt{\frac{8(0.30)}{9.8}} = 0.495$ s. At $240$ frames per second the jump occupies $(0.495)(240) = 119$ frames, which is why a high speed camera is needed: an ordinary $30$ frames per second video would show the whole jump in about $15$ images, far too few to locate the apex.

Biological aside worth a sentence in a laboratory report: the animal is $3.0$ cm long and clears $0.90$ m, that is $30$ body lengths, and it must reach $3.03$ m/s in the few milliseconds during which its legs are still in contact with the ground.

d)

The range on level ground is $R = \frac{v_{0}^{2}\sin 2\theta}{g}$ , and the Moon changes only $g$ . The same legs give the same take off velocity, so the range is INVERSELY proportional to $g$ : dividing $g$ by six multiplies the range by six, giving about $5.4$ m. The flight time is multiplied by six as well, and the maximum height too, so the whole jump is the same shape simply stretched.

e)

Sketch first, origin at the take off point, $x$ horizontal and downhill, $y$ up. The slope is the line $y = -x\tan\alpha$ , and that is the insight the whole part rests on: at landing the drop is tied to the horizontal distance, so the landing condition links the two columns.

Horizontal: $x = v_{0}\cos\theta\,t$ . Vertical: $y = v_{0}\sin\theta\,t - \frac{1}{2}gt^{2}$ .

Landing, $y = -x\tan\alpha$ : $v_{0}\sin\theta\,t - \frac{1}{2}gt^{2} = -v_{0}\cos\theta\tan\alpha\,t$ .

Dividing by $t$ , which discards the root $t = 0$ at take off, gives $\frac{1}{2}gt = v_{0}(\sin\theta + \cos\theta\tan\alpha)$ , so

\displaystyle t = \frac{2v_{0}(\sin\theta + \cos\theta\tan\alpha)}{g}

Checks. Unit: $\frac{\text{m/s}}{\text{m/s}^{2}} = \text{s}$ .

Case $\alpha = 0$ : $\tan\alpha = 0$ and the level ground time $\frac{2v_{0}\sin\theta}{g}$ comes back. The extra term is the time the slope gives away by falling away under the jump.

Sense of variation: $\alpha$ enters only through $\tan\alpha$ , which grows with the slope, so a steeper slope gives a longer flight, as it must, since the ground drops away faster under the animal. At $30^{\circ}$ the flight is $1.146/0.80 = 1.43$ times the level ground one.

And a harder jump, larger $v_{0}$ , lengthens the flight in proportion.

Numbers, with $\sin\theta = 0.80$ , $\cos\theta = 0.60$ , $\tan 30^{\circ} = 0.577$ and $v_{0} = 3.03$ m/s:

$t = \frac{2(3.03)(0.80 + 0.346)}{9.8} = 0.709$ s, against $0.495$ s on the flat.

Horizontal distance $x = (3.03)(0.60)(0.709) = 1.29$ m, and along the slope $\frac{x}{\cos 30^{\circ}} = \frac{1.29}{0.866} = 1.49$ m.

The trap and its price. Two of them here. First, measuring $H$ and $R$ and then computing the angle as $\arctan\frac{H}{R}$ , without the factor $4$ : that gives $18.4^{\circ}$ instead of $53.1^{\circ}$ , and every subsequent answer is wrong. The factor $4$ is not decoration, it is the ratio between the apex height and the height the straight line would have at mid range. Second, dividing the range by the flight time and calling the result the take off speed: that quotient is $\frac{0.90}{0.495} = 1.82$ m/s, which is the HORIZONTAL component and nothing else. The take off speed is the length of a vector, never the length of one of its shadows.

e) The same jump launched from a

30^{\circ}

slope: the ground falls away under the parabola, the flight lasts

0.709

s instead of

0.495

s, and the landing point is

1.49

m down the slope.

Exercise 10: A supply drop, seen from the ground and from the cockpit

A helicopter flies horizontally at $40$ m/s, at a constant height of $44.1$ m, to drop a sealed box of vaccines at a clinic. The box is released without any push, so it leaves the aircraft with the velocity the aircraft has. Take $g = 9.8$ m/s $^{2}$ and keep ignoring the air.

This last exercise puts the two halves of the chapter in the same picture: a projectile for the observer on the ground, a simple vertical fall for the pilot.

a) How long does the box fall?
b) How far ahead of the clinic must the release happen?
c) Give the speed and the direction of the box at impact.
d) The helicopter keeps flying straight at $40$ m/s. Where is the box, relative to the aircraft, at the moment it lands? Describe the fall as the pilot sees it.
e) A second drop is made from $88.2$ m, twice as high, at the same speed. Is the horizontal distance doubled?
f) On a second pass, a crew member shoves the box out horizontally, towards the TAIL, at $25$ m/s relative to the helicopter, just as the aircraft passes directly over the clinic. How far from the clinic, and on which side, does the box land? How far apart are the box and the helicopter at that moment, and what path does the pilot see?

Type your answers, the page tells you right or wrong 0/10

a)

Fall time s

b)

Release distance before the clinic m

c)

Speed at impact m/s

Angle below the horizontal °

d)

Where is the box when it lands?

e)

Is the distance doubled?

New horizontal distance m

f)

Distance from the clinic m

Side of the clinic

Gap between box and helicopter m

Show the solution

Answers

a) $t = 3.0$ s
b) $120$ m before the clinic
c) $v = 49.6$ m/s, at $36.3^{\circ}$ below the horizontal
d) Directly below the helicopter. In the cockpit frame the box falls in a straight vertical line
e) No: the distance becomes $170$ m, multiplied by $\sqrt{2}$ and not by $2$
f) $45$ m past the clinic, on the side the helicopter is heading; the helicopter is then $75$ m further on, and the pilot sees the box drop back along a half parabola ending $75$ m behind

Two columns again. Horizontal: $v_{x} = 40$ m/s, constant, because the box keeps whatever horizontal velocity it had at release and nothing acts sideways. Vertical: $v_{0y} = 0$ , since the release adds no push, and $a_{y} = g$ downward over $44.1$ m.

a)

Vertical column alone: $44.1 = \frac{1}{2}(9.8)t^{2}$ , so $t^{2} = 9.0$ and $t = 3.0$ s. The $40$ m/s plays no part whatsoever, which is the same statement as exercise 1 made with a bench and a marble.

b)

The time crosses over: $x = (40)(3.0) = 120$ m. The pilot must therefore press the button $120$ m BEFORE the clinic, not above it. Releasing overhead puts the box $120$ m past the target, and $120$ m is a serious miss for a medical drop.

c)

At impact $v_{x} = 40$ m/s and $v_{y} = gt = (9.8)(3.0) = 29.4$ m/s downward. Speed: $v = \sqrt{40^{2} + 29.4^{2}} = \sqrt{2464} = 49.6$ m/s, arriving at $\arctan\frac{29.4}{40} = 36.3^{\circ}$ below the horizontal. That is roughly $180$ km/h, which is why such a package needs a parachute in real life, and why the course insists it is ignoring the air.

d)

During those $3.0$ s the box advances $120$ m horizontally, and so does the helicopter, since both keep the same constant horizontal velocity. The box therefore lands exactly below the aircraft. In the frame of the pilot, whose frame moves at constant velocity, the box has no horizontal velocity at all: it simply drops in a straight vertical line and stays under the cockpit the whole way. Same event, two descriptions, and the relative velocity rule of exercise 5 is what connects them: $\vec{v}_{\text{box}/\text{helicopter}} = \vec{v}_{\text{box}/\text{ground}} - \vec{v}_{\text{helicopter}/\text{ground}}$ , whose horizontal part is $40 - 40 = 0$ at every instant.

Worth adding, because examiners ask it: if the helicopter accelerated or turned after the release, the box would no longer stay below it. The two only share a path while they share a velocity, and the box, once released, keeps its own for ever.

e)

No. The height sits under a square root: $t = \sqrt{\frac{2h}{g}}$ , so doubling $h$ multiplies the time, and therefore the horizontal distance, by $\sqrt{2}$ . Here $t = \sqrt{18} = 4.24$ s and $x = (40)(4.24) = 170$ m, not $240$ m. To double the horizontal distance one would have to fly four times higher.

f)

Velocities first, in the ground frame: $\vec{v}_{\text{box}/\text{ground}} = \vec{v}_{\text{box}/\text{helicopter}} + \vec{v}_{\text{helicopter}/\text{ground}}$ , so the horizontal part is $40 - 25 = 15$ m/s, still forward.

The fall still lasts $3.0$ s: the shove is horizontal and the vertical column is exactly that of part a).

Ground view: $x = (15)(3.0) = 45$ m past the clinic, along a flatter half parabola than the one of part b).

The helicopter covers $(40)(3.0) = 120$ m, so it is $120 - 45 = 75$ m ahead of the box when the box lands.

Pilot's view: the box moves back at $25$ m/s relative to the cockpit while it falls, a half parabola ending $(25)(3.0) = 75$ m behind, the same $75$ m measured from the other end.

Pitfall: placing the box $(25)(3.0) = 75$ m BEFORE the clinic, as if the shove replaced the aircraft's velocity instead of adding to it; only a shove of the full $40$ m/s would cancel it.

The trap and its price. The dangerous answer to part b) is to release the box overhead, and the dangerous reasoning behind it is that a dropped object falls straight down. It does, in the frame of whoever is moving with it, and only there. On the ground the box describes a half parabola $120$ m long. The same confusion costs part d) when a student answers that the box lands far behind the helicopter, forgetting that nothing slows it horizontally once the air is ignored.

Corrected exercises: projectile motion and relative velocity (PHYS 101)

Course recap

Part A: the basics (/50)

Exercise 1: Two columns and one clock

Exercise 2: Launched at an angle, and the apex that is not a stop

Exercise 3: A strobe photograph, read column by column

Exercise 4: Two angles, one range

Exercise 5: Relative velocity, from a bush plane in a crosswind to a rainy windshield

Part B: problems and reasoning (/50)

Exercise 6: Thrown from a cliff, where the symmetry is gone

Exercise 7: Crossing a river, and the heading that cancels the current

Exercise 8: Five statements to correct, then a multiple choice question

Exercise 9: A grasshopper filmed at 240 frames per second

Exercise 10: A supply drop, seen from the ground and from the cockpit

See also

Stuck on PHYS 101 projectile problems?