A ball is projected vertically down with an initial velocity from a height of 20 m onto a horizontal floor. During the impact it loses 50% of its energy and rebounds to the same height. The initial velocity of its projection is (in m/s)  [Take g = 10 m/s²]

Let ball is projected vertically downward with velocity v from height h
Total energy at point  $\mathrm{A}=\frac{1}{2}{\mathrm{mv}}^2+\mathrm{mgh}$

During collision loss of energy is 50% and the ball rises up to same height. It means it possess only potential energy at same level.

$50\mathrm{\%} \left(\frac{1}{2}{\mathrm{mv}}^2+\mathrm{mgh}\right)=\mathrm{mgh}$

$\frac{1}{2}\left(\frac{1}{2}{\mathrm{mv}}^2+\mathrm{mgh}\right)=\mathrm{mgh}$

$\mathrm{v}=\sqrt{2\mathrm{gh}}=\sqrt{2\times 10\times 20}$

$\therefore \mathrm{v}=20\mathrm{m}/\mathrm{s}$