Kinematics: Question 3

Worked solution

Setting up a sign convention

Take upwards as positive. The stone is thrown with $u = +12.0\text{ m s}^{-1}$, and the acceleration due to gravity acts downwards, so $a = -9.81\text{ m s}^{-2}$ throughout the motion (both on the way up and on the way down).

Part (a): Maximum height reached

At the maximum height, the stone is momentarily at rest, so $v = 0$. Using $v^2 = u^2 + 2as$ with $s = h_{max}$: $0 = (12.0)^2 + 2(-9.81)h_{max}$ $0 = 144 - 19.62\,h_{max}$ $h_{max} = \frac{144}{19.62}$ $h_{max} = 7.34\text{ m } (3\text{ s.f.})$

Part (b): Time to return to the release height

Method 1 (symmetry): The time to rise to maximum height equals the time to fall back down to the same height. Using $v = u+at$ with $v=0$: $0 = 12.0 - 9.81\,t_{up} \implies t_{up} = \frac{12.0}{9.81} = 1.223\text{ s}$ The total time to return to the release height is twice this: $t = 2 \times 1.223 = 2.45\text{ s } (3\text{ s.f.})$

Method 2 (check): returning to the same height means the net displacement is zero, $s=0$. Using $s = ut + \tfrac{1}{2}at^2$: $0 = 12.0t - \tfrac{1}{2}(9.81)t^2 \implies t\left(12.0 - 4.905t\right) = 0$ The non-zero solution gives $t = 12.0/4.905 = 2.45\text{ s}$, confirming the answer.

Part (c): Speed hitting the ground

The release point is $18.0\text{ m}$ above the ground, so the ground is $18.0\text{ m}$ below the release point. Taking the net displacement from the release point to the ground as $s = -18.0\text{ m}$ (still with upwards positive), and using the same $u$ and $a$ as before: $v^2 = u^2 + 2as = (12.0)^2 + 2(-9.81)(-18.0)$ $v^2 = 144 + 353.16 = 497.16$ $v = \sqrt{497.16} = 22.3\text{ m s}^{-1} (3\text{ s.f.})$

This is a speed (positive square root); the actual velocity at that instant is directed downwards, since the stone is falling when it lands. Note this equation applies directly whether the stone goes up first or not, because suvat only depends on the net displacement, not the path taken.

Final answers

• (a) Maximum height above the release point $= \boxed{7.34}\text{ m}$
• (b) Total time to return to release height $= \boxed{2.45}\text{ s}$
• (c) Speed on hitting the ground $= \boxed{22.3}\text{ m s}^{-1}$