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

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.

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

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.

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.

They land together, but NOT at the same speed, and the two answers do not contradict each other. The vertical components are equal, $4.9$ m/s downward for both, since part d) showed that their vertical columns are identical. The dropped ball has nothing else, so its speed is $4.9$ m/s. The rolled ball still carries its $4.0$ m/s sideways, so its speed is the $6.3$ m/s of part c). There IS enough information, and the reason is worth writing: a speed is the length of the whole vector, $\sqrt{v_{x}^{2} + v_{y}^{2}}$, and the rolled ball has one more nonzero component. The energy route gives the same verdict in one line, $v^{2} = v_{0}^{2} + 2gh$, that is $0 + 24.01$ against $16 + 24.01$.

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.

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.

$H = \frac{v_{0y}^{2}}{2g} = \frac{15^{2}}{2(9.8)} = \frac{225}{19.6} = 11.5$ 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.

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.

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.

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.

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.

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.

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.

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.

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.

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 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.

Yes, and part c) already contains the whole answer. The distance is the $35.3$ m of part a), so the two twin angles $30^{\circ}$ and $60^{\circ}$ both land on the friend. They do not take the same time: $3.53$ s for the steep shot, $2.04$ s for the flat one. For both balloons to ARRIVE together, the slow one must LEAVE first: fire at $60^{\circ}$, swing the launcher down to $30^{\circ}$, and fire again $3.53 - 2.04 = 1.49$ s later. The steep balloon is then still climbing when the flat one leaves, and both land at $t = 3.53$ s after the first shot.

The trick rests on having TWO angles for one range, and that fails at both ends. At $40.8$ m, the maximum range, the two twin angles merge into one, $45^{\circ}$, so there is a single flight time and nothing to stagger: two balloons fired at different instants land at different instants. Beyond $40.8$ m, part d) showed that no angle reaches at all. The hint of the tutorial, that the range is the same for both balloons, is exactly what points to the twin angles; the question to ask next is always whether two such angles still exist.

Sketch first, axes written on the sketch: $x$ east, $y$ north. Then the chain, read out loud: goose relative to ground equals goose relative to air plus air relative to ground, $\vec{v}_{G/g} = \vec{v}_{G/a} + \vec{v}_{a/g}$. In components, $\vec{v}_{G/a} = (0\,;-25)$ and $\vec{v}_{a/g} = (8.0\,;0)$, so $\vec{v}_{G/g} = (8.0\,;-25)$ m/s. Magnitude: $\sqrt{8.0^{2} + 25^{2}} = \sqrt{689} = 26$ m/s. Direction: $\arctan\frac{8.0}{25} = 17.7^{\circ}$ east of south. The goose points south and drifts east, exactly like the boat of exercise 7 whose bow points across the river; pointing somewhere is not the same as going there.

Same chain, but now the unknown is the direction of $\vec{v}_{G/a}$, whose LENGTH is fixed at $25$ m/s, while $\vec{v}_{G/g}$ has a known direction, due south, and an unknown length. Two unknowns, and a vector equation gives two component equations, so it can be solved. East component: $v_{G/a,x} + 8.0 = 0$, so the goose must fly $8.0$ m/s WEST relative to the air, that is into the wind. The heading therefore satisfies $\sin\alpha = \frac{8.0}{25}$, $\alpha = 18.7^{\circ}$ west of south. North component: $v_{G/g} = 25\cos\alpha = \sqrt{25^{2} - 8.0^{2}} = \sqrt{561} = 23.7$ m/s due south. The geometric route gives the same numbers from the tip to tail drawing: a right triangle whose HYPOTENUSE is the $25$ m/s of the goose relative to the air, one leg the $8.0$ m/s of wind, the other the ground speed. Compare with part a), where the $25$ m/s was a leg and the ground speed the hypotenuse; mixing up the two triangles is the classic error here, and it gives $26$ m/s in part b) where the answer must be SMALLER than $25$ m/s, since part of the effort is spent cancelling the wind.

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 $\vec{v}_{\text{rain}/\text{car}} = \vec{v}_{\text{rain}/\text{ground}} - \vec{v}_{\text{car}/\text{ground}} = (0 - 15\,;-8.0 - 0) = (-15\,;-8.0)$.

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 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.

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.

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

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.

From ground level the symmetric formula is allowed: $d = \frac{v_{0}^{2}\sin 2\theta}{g}$, so $v_{0} = \sqrt{\frac{g\,d}{\sin 82^{\circ}}} = \sqrt{\frac{(9.8)(84.8)}{0.990}} = 29.0$ m/s. Prediction for the real throw: a ball released $1.6$ m up has to fall $1.6$ m further than it rose, so it stays in the air LONGER than the same launch from the ground. To cover the same $84.8$ m in more time it needs less horizontal speed, hence a SMALLER launch speed. Writing this prediction before the algebra is what lets you catch a sign error in part f).

Same two columns as parts a) to d), origin at the hand, up positive, so the ground is at $y = -h$. Horizontal column: $d = v_{0}\cos\theta\,t$, hence $t = \frac{d}{v_{0}\cos\theta}$. Vertical column at landing: $-h = v_{0}\sin\theta\,t - \frac{1}{2}gt^{2}$. Carrying $t$ across gives $-h = d\tan\theta - \frac{g\,d^{2}}{2v_{0}^{2}\cos^{2}\theta}$, and solving for the only unknown: $v_{0}^{2} = \frac{g\,d^{2}}{2\cos^{2}\theta\,(d\tan\theta + h)}$. Here two equations and two unknowns, $v_{0}$ and $t$, is what makes the problem solvable at all.

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.

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.

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.

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.