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) $14\ \text{m s}^{-1}$
Concept. Energy just before impact is twice the energy after (which gives a $10$ m rise).
Step 1. Energy before impact. $mg(20)$.
Step 2. At 10 m. $\dfrac12v_0^2 + 100 = 200 \Rightarrow v_0 = \sqrt{200} \approx 14$ m/s.
Answer. $14\ \text{m s}^{-1}$ (Option C).
Solution by Sreeraj P, M.Sc Physics