Question

A ball is dropped vertically from rest at a height of 12 m. After striking the ground, it bounces to a height of 9 m. What fraction of kinetic energy does it loose on striking the ground?

Answer 1

$$\begin{gathered} \mathrm{Fractional}\mathrm{loss}\mathrm{of}\mathrm{KE}=\mathrm{Fractional}\mathrm{loss}\mathrm{of}\mathrm{PE} \\ =\frac{\mathrm{mg}\left({\mathrm{h}}_1-{\mathrm{h}}_2\right)}{{\mathrm{mgh}}_1} \\ =\frac{{\mathrm{h}}_1-{\mathrm{h}}_2}{{\mathrm{h}}_1} \\ =\frac{12-9}{12} \\ =0.25\mathrm{or}25\% \end{gathered}$$