Q 11-05-028NEETJEE MainEAMCET 2010 (Engineering)Medium
A ball is falling freely from a certain height. When it reaches $10$ m height from the ground its velocity is $v_0$. It collides with the horizontal ground and loses $50\%$ of its energy and rises back to a height of $10$ m. The value of $v_0$ is ($g = 10\ \text{m/s}^2$)
Answer: (C) $14\ \text{m s}^{-1}$
After the bounce, the energy left is $mg(10)$, which is half the energy just before impact. So just before impact the energy was $mg(20)$.
At $10$ m height the energy is the same:
$$\frac{1}{2}mv_0^2 + mg(10) = mg(20) \;\Rightarrow\; v_0^2 = 200 \;\Rightarrow\; v_0 \approx 14\ \text{m s}^{-1}$$
Solution by Sreeraj P, M.Sc Physics