Kinematics of Motion in a Straight Line: Question 4

Worked solution

Part (a): Greatest height

Taking upward as positive, $u = 25\text{ m s}^{-1}$ and $a = -g = -10\text{ m s}^{-2}$ (gravity acts downward). At the greatest height, the stone is instantaneously at rest, so $v=0$.

Using $v^2 = u^2 + 2as$:

$0 = 25^2 + 2(-10)s = 625 - 20s$

$20s = 625 \implies s = 31.25$

So the greatest height is $31.25$ m.

Recompute independently as a check: first find the time to reach the top using $v=u+at$: $0 = 25 - 10t \implies t = 2.5$ s. Then find the height using $s=ut+\tfrac12at^2$: $s = 25(2.5) - \tfrac12(10)(2.5)^2 = 62.5 - 31.25 = 31.25$. This matches, confirming the greatest height is $31.25$ m.

Part (b): Total time to return to the ground

The stone returns to the ground when its displacement from the starting point is zero, i.e. $s=0$. Using $s = ut + \tfrac12 at^2$:

$0 = 25t - \frac12(10)t^2 = 25t - 5t^2$

$0 = 5t(5-t)$

So $t=0$ (the start) or $t=5$. The total time for the stone to return to the ground is $5$ s.

Recompute independently as a check: by symmetry of motion under constant gravity, the time to fall back down from the greatest height equals the time taken to rise, i.e. $2.5$ s (found in part (a)). So the total flight time is $2.5+2.5=5$ s, which matches.

Part (c): Speed at landing

Using $v = u + at$ with $t=5$:

$v = 25 + (-10)(5) = 25 - 50 = -25$

The negative sign shows the velocity is directed downward (opposite to the initial throw). The speed is the magnitude of this velocity, so the stone lands at a speed of $25\text{ m s}^{-1}$.

Recompute independently as a check: using $v^2 = u^2+2as$ with $s=0$ (back at the starting height): $v^2 = 25^2 + 2(-10)(0) = 625$, so $v=\pm 25$. Since the stone is moving downward at landing, $v=-25\text{ m s}^{-1}$, giving a speed of $25\text{ m s}^{-1}$. Matching the launch speed exactly, as expected since there is no air resistance.

Final answers

• (a) Greatest height $=\boxed{31.25\text{ m}}$
• (b) Total time to return to the ground $=\boxed{5\text{ s}}$
• (c) Speed at landing $=\boxed{25\text{ m s}^{-1}}$