Introduction to Projectile Motion — Practice Problem Solutions

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

1. $t=\sqrt{\dfrac{2(4.9)}{9.8}}=\sqrt{1}=1\text{ s}$. Horizontal distance $=(3)(1)=3\text{ m}$.

2. $t=\sqrt{\dfrac{2(19.6)}{9.8}}=\sqrt{4}=2\text{ s}$. Horizontal distance $=(5)(2)=10\text{ m}$.

3. $t=\sqrt{\dfrac{2(44.1)}{9.8}}=\sqrt{9}=3\text{ s}$. Horizontal distance $=(10)(3)=30\text{ m}$.

4. $t=\sqrt{\dfrac{2(19.6)}{9.8}}=2\text{ s}$. $v_x=\dfrac{24}{2}=12\text{ m/s}$.

5. $t=\sqrt{\dfrac{2(78.4)}{9.8}}=\sqrt{16}=4\text{ s}$. (a) $v_y=gt=(9.8)(4)=39.2\text{ m/s}$. (b) $v=\sqrt{v_x^2+v_y^2}=\sqrt{15^2+39.2^2}=\sqrt{1761.64}\approx42.0\text{ m/s}$. (c) $\theta=\arctan\left(\dfrac{39.2}{15}\right)\approx69.1°$ below horizontal.

6. a) $t=\dfrac{2\text{ m}}{4\text{ m/s}}=0.5\text{ s}$. b) Vertical drop $=\tfrac{1}{2}gt^2=\tfrac{1}{2}(9.8)(0.5)^2=1.225\text{ m}$ — exactly equal to the dart's throw height. That means the dart has fallen all the way to the ground by the time it reaches the target's horizontal position; it does not hit the bullseye, landing instead at the base of the target, 1.225 m below where it was aimed. c) The thrower should increase the horizontal velocity. A faster dart reaches the target in less time, giving gravity less time to pull it downward, so it strikes higher up — closer to the bullseye at the original throw height.

7. False. Horizontal and vertical motion are independent in projectile motion — the horizontal velocity has no effect on how quickly gravity pulls the bullet down. Both the fired bullet and the dropped bullet start their fall from the same height with the same (zero) initial vertical velocity, so they experience identical vertical motion and hit the ground at exactly the same time, regardless of how far the fired bullet travels horizontally.