A boy finds himself at the bottom of a hole 1.25 meters deep. If the boy is to jump out of the hole, with what minimum velocity must he launch himself?

Nov 23, 2015

$4.95 \text{m/s}$

Explanation:

${v}^{2} = {u}^{2} + 2 a s$

In this case we can set $v = 0$ which means he will just reach the surface:

$\therefore 0 = {u}^{2} - 2 \times g \times 1.25$

${u}^{2} = 2 \times 9.9 \times 1.25$

$u = 4.95 \text{m/s}$