Question

A diver of mass 65kg takes off from the 10m platform from a standing position by jumping upwards with a vertical spread of 2.0m/s .Calculate
-time taken to reach the water
-gravitational potential energy
-kinetic energy

Answer 1

$a) h=\frac{gt^2}{2}-v_0t\\ \frac{gt^2}{2}-v_0t-h=0\\ t=\frac{v_0\pm\sqrt{v_0^2+2gh}}{g} = \frac{2\pm\sqrt{4+2\cdot9.81\cdot10}}{9.81}=\left[\begin{array}{ccc}1.65\\-1.24\end{array}\right.$

time taken to reach the water is equal to 1.65s

b) $E_p = mgh = 65\cdot9.81\cdot10 = 6376.5 J$

c) $E_{kin} = \frac{mv^2_0}{2}+E_p = \frac{65\cdot2^2}{2}+6376.5 = 6505.5 J$