Projectile Motion

A projectile moves through the air under gravity alone, with no thrust or driving force. Once the netball leaves the hand only gravity acts on it, so it is a projectile. The rocket and car have their own propulsion, so they are not.

Horizontal and vertical motions are independent. Both bullets start with zero vertical velocity and fall under the same acceleration $g$, so their vertical motions are identical and they land together, whatever the horizontal speed.

At the top the vertical velocity is momentarily zero, so the velocity is purely horizontal (equal to the launch horizontal component). But gravity never switches off: the acceleration is still $9.8\text{ m/s}^2$ downward throughout. Zero velocity does not mean zero acceleration.

Range depends on $\sin(2\theta)$, which is the same for complementary angles (those adding to $90^\circ$). Since $30^\circ + 60^\circ = 90^\circ$, the two give equal ranges (the steeper one just flies higher for longer).

The vertical component uses sine: $u_y = v\sin\theta = 20 \times 0.6 = 12\text{ m/s}$. (The horizontal component uses cosine: $u_x = 20 \times 0.8 = 16\text{ m/s}$.)

At the top the vertical velocity is zero, so $h = \dfrac{u_y^2}{2g} = \dfrac{12^2}{2(9.8)} = \dfrac{144}{19.6} \approx 7.3\text{ m}$. Only the vertical component sets the height.

Flight time $t = \dfrac{2u_y}{g} = \dfrac{2(12)}{9.8} \approx 2.45\text{ s}$; then $x = u_x t = 16 \times 2.45 \approx 39\text{ m}$. Use the full flight time, not the time to the top.

$\tan\theta = \dfrac{4h}{R} = \dfrac{4(15)}{80} = \dfrac{60}{80} = 0.75$, so $\theta = \tan^{-1}(0.75) \approx 37^\circ$. (This handy relation comes from dividing $h = \frac{u_y^2}{2g}$ by $R = \frac{2u_x u_y}{g}$.)

$14.7 = 19.6t - 4.9t^2$, so $4.9t^2 - 19.6t + 14.7 = 0$, i.e. $t^2 - 4t + 3 = 0$ and $(t-1)(t-3) = 0$. The projectile passes that height twice: at $t = 1\text{ s}$ going up and $t = 3\text{ s}$ coming down.

Energy conservation gives $\tfrac{1}{2}v^2 = \tfrac{1}{2}u^2 - gh$, so $v^2 = 25^2 - 2(9.8)(10) = 625 - 196 = 429$ and $v \approx 20.7\text{ m/s}$. The answer does not depend on the launch angle, only on the height.

The range is where the path returns to the ground, $y = 0$: $x(0.75 - 0.0125x) = 0$, giving $x = 0$ (launch) or $x = \dfrac{0.75}{0.0125} = 60\text{ m}$. (The coefficient of $x$ also tells you $\tan\theta = 0.75$, a $37^\circ$ launch.)

Landing at $45^\circ$ means the vertical speed equals the (constant) horizontal speed: $v_y = v_x = 15\text{ m/s}$. Then $t = \dfrac{v_y}{g} = 1.53\text{ s}$ and $h = \tfrac{1}{2}gt^2 = \tfrac{1}{2}(9.8)(1.53)^2 \approx 11.5\text{ m}$.

$R = \dfrac{v^2\sin(2\theta)}{g}$, so $\sin(2\theta) = \dfrac{Rg}{v^2} = \dfrac{35(9.8)}{400} \approx 0.858$. Then $2\theta \approx 59^\circ$ or $121^\circ$, giving $\theta \approx 30^\circ$ or $60^\circ$ (a low, fast shot or a high, lobbed one).

Time to reach the wall: $t = \dfrac{x}{u_x} = \dfrac{20}{15} = 1.33\text{ s}$. Height then: $y = u_y t - \tfrac{1}{2}gt^2 = 20(1.33) - 4.9(1.33)^2 \approx 26.7 - 8.7 = 18\text{ m}$. It clears the $6\text{ m}$ wall comfortably.

The clearance is the ball's height at the wall minus the wall height: $18 - 6 = 12\text{ m}$. (A clearance question is really just "find $y$ at that $x$", then subtract the obstacle height.)

$R = \dfrac{v^2\sin(2\theta)}{g} = \dfrac{25^2\sin 60^\circ}{9.8} = \dfrac{625(0.866)}{9.8} \approx 55\text{ m}$. Since $55 > 40$, the rider clears the gap with about $15\text{ m}$ to spare.

$45^\circ$ maximises range only when launch and landing are at the same height. Launching from a height means extra fall time is available, so a flatter shot (more horizontal speed, angle $<45^\circ$) travels further. The higher the cliff, the lower the optimal angle.

Take up as positive, ground $25\text{ m}$ below launch: $-25 = 12t - 4.9t^2$, i.e. $4.9t^2 - 12t - 25 = 0$. The quadratic formula gives $t = \dfrac{12 + \sqrt{144 + 490}}{9.8} \approx 3.8\text{ s}$ (positive root).

Horizontal velocity is constant, so $x = u_x t = 16 \times 3.8 \approx 61\text{ m}$. Once you have the flight time from the vertical quadratic, the horizontal distance is a simple product.

Vertical velocity at landing: $v_y = u_y - gt = 12 - 9.8(3.8) \approx -25.2\text{ m/s}$. With $v_x = 16\text{ m/s}$: $v = \sqrt{16^2 + 25.2^2} = \sqrt{891} \approx 30\text{ m/s}$.

The horizontal velocity is constant, so read it off the instant given: $v_x = 20\cos 37^\circ = 16\text{ m/s}$. At launch that same $v_x = v\cos 53^\circ$, so $v = \dfrac{16}{\cos 53^\circ} = \dfrac{16}{0.6} \approx 27\text{ m/s}$.

Without gravity the dart would travel straight to the monkey's original spot. Gravity makes both the dart and the monkey fall by the same $\tfrac{1}{2}gt^2$ in the same time, so the dart drops exactly as much as the monkey and still hits it. (Aiming straight at the target works.)

First the angle: $\tan\theta = \dfrac{4h}{R} = \dfrac{80}{60} = 1.33$, so $\theta \approx 53^\circ$. Then $R = \dfrac{v^2\sin(2\theta)}{g}$ gives $v^2 = \dfrac{Rg}{\sin 106^\circ} = \dfrac{60(9.8)}{0.96} \approx 613$, so $v \approx 25\text{ m/s}$.

Flight time from the drop: $t = \sqrt{\dfrac{2h}{g}} = \sqrt{\dfrac{2(30)}{9.8}} \approx 2.47\text{ s}$. To cover $40\text{ m}$ horizontally in that time: $v = \dfrac{x}{t} = \dfrac{40}{2.47} \approx 16\text{ m/s}$.

Energy conservation from launch to the ground $20\text{ m}$ lower: $\tfrac{1}{2}v^2 = \tfrac{1}{2}u^2 + gh$, so $v^2 = 25^2 + 2(9.8)(20) = 625 + 392 = 1017$ and $v \approx 32\text{ m/s}$. The launch angle does not matter, only the drop in height, which makes energy the quickest route.