Question

A 75 kg swimmer steps off a 10.0 m tower.  What is the swimmer’s velocity on hitting the water?

Answer 1

Given data:

Height of the tower;

$h=10\text{ m}$

The velocity is given as,

$v^2=u^2+2gh$

Here, u is the initial velocity of the swimmer (u=0), and g is the acceleration due to gravity (g=9.8 m/s²).

Substituting all known values,

$\begin{gathered} v^2=0^2+2\times(9.8\text{ m/s}^2)\times(10\text{ m}) \\ v^2=196\text{ m}^2\text{ /s}^2 \\ v=14\text{ m/s} \end{gathered}$

Therefore, the swimmer's velocity on hitting the water is 14 m/s.