1. Tardigrade
  2. Question
  3. Physics
  4. 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

Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

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

Work, Energy and Power

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}H\right) $ .
$ \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} = 2g\left(H-h\right) $
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}m2g\left(H-h\right)}{mgh} = \left(\frac{H}{h}-1\right)= \frac{1}{3} $