A ball is thrown vertically downwards from a height of 20 m with an initial velocity ${\nu }_0$. It collides with the ground, loses 50 percent of its energy in collision and rebounds to the same height. The initial velocity ${\mathrm{\nu }}_0$ is (Take g = $10 {\mathrm{ms}}^{-2}$ )

Let the energy of the ball just before hitting
the ground be E₁.

${\mathrm{E}}_1=\frac{1}{2}{\mathrm{mv}}_0^2+\mathrm{mgh}$

( h is the height of the tower)
after colliding with ground it rebounces with
an energy E₂.

and the ball reaches same height E₂=mgh

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

$\Rightarrow \mathrm{m}\left(\mathrm{g}\right)\left(20\right)=\frac{1}{2}\left(\frac{1}{2}{\mathrm{mv}}_0^2+\mathrm{mg}\left(20\right)\right)$

$\Rightarrow v_0^2=400\Rightarrow v_0=20\mathrm{m}/\mathrm{s}$