Question

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. $5\, ms^{-1}$

Answer 1

Answer: (A)

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

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.
$50 \%\left(\frac{1}{2} m v^{2} +m g h\right)=m g h$
$\frac{1}{2}\left(\frac{1}{2} m v^{2}+m g h\right)=m g h$
$v=\sqrt{2 g h}=\sqrt{2 \times 10 \times 20}$
$\therefore v=20\, m/s$