## In This post you will get answer of Q.

A ball of mass m is dropped from a cliff of height $ H $ . The ratio of its kinetic energy to the potential energy when it is fallen through a height $ 3/4, H $ is

Solution:

## Total mechanical energy at height, $ H $

$ E_H = mgH $

Let $ v_h $ be velocity of the ball at height $ h $ $ left(= frac{3}{4}Hright) $ .

$ therefore $ Total mechanical energy at height $ h $ ,

$ E_{h} =mgh + frac{1}{2}mv^{2}_{h} $

According to law of conservation of mechanical energy,

$ E_{H} = E_{h}; mgH = mgh +frac{1}{2}mv^{2}_{h} $

$ v_{h},^{2} = 2gleft(H-hright) $

Required ratio of kinetic energy to potential energy at height h is

$ frac{K_{h}}{V_{h}} = frac{frac{1}{2}mv^{2}_{h}}{mgh} $

$ = frac{frac{1}{2}m2gleft(H-hright)}{mgh} = left(frac{H}{h}-1right)= frac{1}{3} $