A ball is thrown from a height of h meter with an initial downward velocity ${\mathrm{v}}_{\mathrm{o}}$. It hits the ground, loses half of its kinetic energy and bounces back to the same height. The value of ${\mathrm{v}}_{\mathrm{o}} \mathrm{is}$ 

Total energy at the time of throwing the ball = mgh + $\frac{1}{2}{{\mathrm{mv}}_0}^2$

Energy after collision with ground = $\frac{1}{2}(\mathrm{mgh}+\frac{1}{2}{{\mathrm{mv}}_0}^2)$

The ball again rises to height h.

$\therefore \frac{1}{2}(\mathrm{mgh} + \frac{1}{2}{{\mathrm{mv}}_0}^2) = \mathrm{mgh}$

or mgh $+ \frac{1}{2}\mathrm{m} {{\mathrm{v}}_0}^2 = 2 \mathrm{mgh}$

or $\frac{1}{2}{{\mathrm{mv}}_0}^2 = \mathrm{mgh} \mathrm{or} {\mathrm{v}}_{\mathrm{o}} = \sqrt{2\mathrm{gh}}$