Q 11-05-017NEETJEE MainEAMCET 2013 (Engineering)Easy
A ball at rest is dropped from a height of $12$ m. If it loses $25\%$ of its kinetic energy on striking the ground and bounces back to a height $h$, the value of $h$ is
Answer: (C) $9$ m
Just before impact its kinetic energy is $mg(12)$. It keeps $75\%$ of this, which becomes potential energy at the top of the bounce:
$$mgh = 0.75 \times mg(12) \;\Rightarrow\; h = 9\ \text{m}$$
Solution by Sreeraj P, M.Sc Physics