Question

A ball of mass $0.2 \,Kg$ falls under gravity from a height of $10 \,m$ with an initial velocity. It collides with the floor and looses $50 \%$ of its energy and then rises back to the same height. The value of its initial velocity is

• A. $14 \,ms ^{-1}$
• B. $7\, ms ^{-1}$
• C. $24\, ms ^{-1}$
• D. $2.5 \,ms ^{-1}$

Answer 1

Answer: (A)

Total energy before collision = kinetic energy $+$ potential energy
$=1 / 2\, mv ^{2}+ mgh$
After collision $=1 / 2\left(1 / 2\, m v^{2}+m g h\right)$
Since the ball rebounds to a height $10\, m $
$m g(10)=1 / 2\left(1 / 2 m v^{2}+m g(10)\right)$
$20\, g-10\, g=1 / 2 v^{2} $
$V^{2}=20\, g=20 \times 9.8=196$
$V=\sqrt{196}=14$