Question

A ball is projected vertically upward with a speed of 50 m/s. Find (a) the maximum height, (b) the time to reach the maximum height, (c) the speed at half the maximum height. Take g = 10 m/s².

Answer 1

Given:
Initial speed of the ball, u = 50 m/s
Acceleration, a = −10 m/s²
At the highest point, velocity v of the ball is 0.

(a) $s=\frac{v^2-u^2}{2a}$
⇒ $s=\frac{\left(0^2-{50}^2\right)}{2\times \left(-10\right)}=125\mathrm{m}$
Maximum height = 125 m

(b) $t=\frac{\left(v-u\right)}{a}=\frac{\left(0-50\right)}{-10}=5\mathrm{s}$

(c) $s=\frac{125}{2}=62.5\mathrm{m}$
From v² − u² = 2as, we have:
$$\begin{gathered} v=\sqrt{\left(u^2+2as\right)} \\ \Rightarrow v=\sqrt{{\left(50\right)}^2+2\times \left(-10\right)\times \left(62.5\right)}=\sqrt{\left(2500-1250\right)} \\ \Rightarrow v=\sqrt{1250}\approx 35\mathrm{m}/\mathrm{s} \end{gathered}$$