Question

4. With what initial velocity will a body moving along a vertical line have to be thrown if after 5 sec it is to be 50 ft above its starting place?

Answer 1

Given:

$t=5\:\rm s$

$d=50\:\rm ft$

The vertical distance

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

$$v_0=\frac{2d+gt^2}{2t}\\ =\frac{2*50+32*5^2}{2*5}=90\:\rm ft/s=27.4\: m/s$$