# Collisions

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# 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]

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Solution

## Let ball is projected vertically downward with velocity v from height hTotal 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.$\frac{1}{2}\left(\frac{1}{2}{\mathrm{mv}}^{2}+\mathrm{mgh}\right)=\mathrm{mgh}$$\mathrm{v}=\sqrt{2\mathrm{gh}}=\sqrt{2×10×20}$$\therefore \mathrm{v}=20\text{\hspace{0.17em}}\mathrm{m}/\mathrm{s}$

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