Q. 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
A) $20\ ms^{- 1}$
B) $15\ ms^{- 1}$
C) $10\ ms^{- 1}$
D) $5\ ms^{- 1}$
Explanation
(a) Let ball is projected vertically downward with velocity $v$ from height $h$
Total energy at point $A = \frac{1}{2}mv^{2} + mgh$
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
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$50\%\left( \frac{1}{2}mv^{2} + mgh \right) = mgh$
$\frac{1}{2}\left( \frac{1}{2}mv^{2} + mgh \right) = mgh$
$v = \sqrt{2gh} = \sqrt{2 \times 10 \times 20}$
$\therefore v = 20\ m/s$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.