Work Energy and Power - Result Question 49
52. A ball is thrown vertically downwards from a height of $20 m$ with an initial velocity $v_0$. It collides with the ground loses 50 percent of its energy in collision and rebounds to the same height. The initial velocity $v_0$ is : [2015 RS] (Take $g=10 ms^{-2}$ )
(a) $20 ms^{-1}$
(c) $10 ms^{-1}$
(b) $28 ms^{-1}$
(d) $14 ms^{-1}$
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Answer:
Correct Answer: 52. (a)
Solution:
- (a) When ball collides with the ground it loses its $50 %$ of energy
$ \therefore \frac{KE_f}{KE_i}=\frac{1}{2} \Rightarrow \frac{\frac{1}{2} mV_f^{2}}{\frac{1}{2} mV_i^{2}}=\frac{1}{2} $
or $\frac{V_f}{V_i}=\frac{1}{\sqrt{2}}$
or, $\frac{\sqrt{2 gh}}{\sqrt{V_0^{2}+2 gh}}=\frac{1}{\sqrt{2}}$
or, $4 gh=V_0^{2}+2 gh$
$\therefore V_0=20 ms^{-1}$
