Example

Question:

A ball is thrown vertically upwards with a velocity of \(20\,\mathrm{m\,s^{-1}}\) from the top of a multistorey building. The height of the point from where the ball is thrown is \(25.0\,\mathrm{m}\) from the ground.
(a) How high will the ball rise?
(b) How long will it be before the ball hits the ground?
Take \(g = 10\,\mathrm{m\,s^{-2}}\).

Solution:

(a) Maximum height rise:
Let the \(y\)-axis be vertically upward with zero at the ground.
\(v_0 = +20\,\mathrm{m\,s^{-1}}\), \(a = -g = -10\,\mathrm{m\,s^{-2}}\), \(v = 0\,\mathrm{m\,s^{-1}}\) at the highest point.
Use equation: \[ v^2 = v_0^2 + 2a(y - y_0) \] \[ 0 = (20)^2 + 2(-10)(y - y_0) \] \[ 0 = 400 - 20(y - y_0) \implies y - y_0 = 20\,\mathrm{m} \] So, the ball rises \(20\,\mathrm{m}\) above the launch point.

(b) Time before ball hits the ground:
First Method:
Split path into two parts: upward (A to B) and downward (B to C).
Upward (to max height): \[ v = v_0 + at \implies 0 = 20 - 10 t_1 \implies t_1 = 2\,\mathrm{s} \] Downward (max height to ground):
From B, the ball falls from \(y_0 = 45\,\mathrm{m}\) (\(25 + 20\)), to ground (\(y = 0\)), with \(v_0 = 0\): \[ y = y_0 + v_0 t + \frac{1}{2} a t^2 \implies 0 = 45 + 0 - 5 t_2^2 \implies t_2^2 = 9 \implies t_2 = 3\,\mathrm{s} \] Therefore, total time before hitting ground: \[ t = t_1 + t_2 = 2 + 3 = 5\,\mathrm{s} \] Second Method:
Consider initial position \(y_0 = 25\,\mathrm{m}\), final \(y = 0\), \(v_0 = 20\,\mathrm{m\,s^{-1}}\), \(a = -10\,\mathrm{m\,s^{-2}}\): \[ y = y_0 + v_0 t + \frac{1}{2} a t^2 \] \[ 0 = 25 + 20t - 5t^2 \rightarrow 5t^2 - 20t - 25 = 0 \] Solve quadratic: \(t = 5\,\mathrm{s}\).
Hence, the ball will hit the ground after 5 seconds. The second method is preferable for problems with constant acceleration, as you need not divide the motion path.