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

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 dropped ball 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.

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 $t$.

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.

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, not a graph of the motion in time. Students read its shape as if the horizontal axis were $t$, then announce that the ball slows down near the top because the dots there are closer together in $y$. The dots are equally spaced in $x$ at every instant, which is precisely the point of the photograph. 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.

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.

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.

a) Both velocities point east, so the vector sum collapses to a subtraction of signed numbers: $v_{A/B} = v_{A/\text{ground}} - v_{B/\text{ground}} = 25 - 18 = 7$ m/s east. In the truck driver's frame, car A creeps up at $7$ m/s and nothing else happens. The gap of $40$ m closes in $t = \frac{40}{7} = 5.7$ s.

b) Now $v_{B/\text{ground}} = -18$ m/s with east positive, so $v_{A/B} = 25 - (-18) = 43$ m/s east. The closing speed adds up in a head on approach, which is why the same $430$ m are eaten in $t = \frac{430}{43} = 10$ s instead of a minute. Same physics, same formula, only the sign of one number changed.

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.

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.

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.

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 $7.0$.

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) TRUE. The vertical column of the two balls is identical: same initial vertical velocity, zero, same acceleration $g$, same height to fall. The rolled ball also travels sideways, and travelling sideways costs no time, because the horizontal motion has no influence on the vertical one. A bench $1.225$ m high gives $t = 0.50$ s for both, whatever the rolling speed. This is the experiment on which the whole chapter rests, and it is worth remembering as a picture rather than a formula.

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: four of them are the same mistake wearing different clothes, treating a quantity of one column as if it belonged to the other, or treating a proportionality as if it were linear when a square is involved. Before accepting any sentence about a projectile, ask which column each quantity lives in, and whether the relation is linear, quadratic or trigonometric.

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.

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.

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.

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.