Question

A ball is thrown vertically downward from a cliff 48 m high and reaches the ground 2 s later. What was the ball's initial velocity ?

Answer 1

Given:

$h=48\:\rm m$

$g=9.8\:\rm m/s$

$t=2\:\rm s$

The distance traveled by a ball

$$d=v_0t+gt^2/2$$

Hence, the initial speed of a ball

$$v_0=d/t-gt/2\\ =48/2-9.8*2/2=14.2\:\rm m/s$$