(1998)
Solution:
The velocity before rebound is $v_1 = \sqrt{2gh_1}$ and after rebound is $v_2 = \sqrt{2gh_2}$. The ratio is $\frac{v_2}{v_1} = \sqrt{\frac{h_2}{h_1}} = \sqrt{\frac{1.8}{5}} = \frac{3}{5}$.
A ball is dropped from a height of $5\text{ m}$, if it rebound upto height of $1.8\text{ m}$, then the ratio of velocities of the ball after and before rebound is:
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