Angled Projectile Motion — Practice Problem Solutions

Using $g=9.8\text{ m/s}^2$.

1. $v_x=14\cos45°\approx9.90\text{ m/s}$; $v_y=14\sin45°\approx9.90\text{ m/s}$.

2. $v_y=20\sin30°=10\text{ m/s}$. $t_{up}=\dfrac{v_y}{g}=\dfrac{10}{9.8}\approx1.02\text{ s}$.

3. $v_x=25\cos40°\approx19.15\text{ m/s}$, $v_y=25\sin40°\approx16.07\text{ m/s}$. (b) Total flight time $=\dfrac{2v_y}{g}\approx3.28\text{ s}$. (c) Max height $=\dfrac{v_y^2}{2g}=\dfrac{16.07^2}{19.6}\approx13.2\text{ m}$.

4. $v_x=18\cos53°\approx10.83\text{ m/s}$, $v_y=18\sin53°\approx14.38\text{ m/s}$. Flight time $=\dfrac{2v_y}{g}\approx2.93\text{ s}$. Range $=v_x\times t\approx(10.83)(2.93)\approx31.8\text{ m}$.

5. $20°$ and $70°$ are complementary angles ($20°+70°=90°$), and range depends on $\sin(2\theta)$, which gives the same value for complementary angles: $\sin(40°)=\sin(140°)$. Both launches land at the same range: $R=\dfrac{v^2\sin(2\theta)}{g}=\dfrac{16^2\sin40°}{9.8}\approx16.8\text{ m}$ — identical ranges, even though one is a low, flat arc and the other is a high, steep one.

6. a) $v_x=30\cos35°\approx24.57\text{ m/s}$, $v_y=30\sin35°\approx17.21\text{ m/s}$. b) $t=\dfrac{2v_y}{g}\approx3.51\text{ s}$. c) $R=v_x\times t\approx(24.57)(3.51)\approx86.3\text{ m}$. d) $55°$ and $35°$ are complementary, so the range stays the same, $\approx86.3\text{ m}$, WITHOUT recalculating. What does change: at $55°$ the launch is steeper, so $v_y$ becomes the larger component ($30\sin55°\approx24.57\text{ m/s}$) while $v_x$ becomes the smaller one ($30\cos55°\approx17.21\text{ m/s}$) — exactly swapped from the $35°$ case. That means the ball reaches a greater maximum height and stays in the air longer at $55°$, even though it travels the same horizontal distance.

7. False. Range depends on $\sin(2\theta)$, which increases only up to $\theta=45°$ (where $2\theta=90°$ and $\sin(2\theta)=1$, its maximum) and then decreases again as the angle keeps increasing past $45°$ toward $90°$. A ball launched straight up at $90°$ has zero range, just like one "launched" at $0°$ — so increasing the angle does not always increase the range; it only helps up to $45°$.