Question

A ball is thrown vertically downwards from a height of 20 m with an initial velocity ${\upsilon }_0$. It collides with the ground, loses 50 percent of its energy in collision and rebounds to the same height. The initial velocity ${\upsilon }_0$ is (Take $g=10m{s}^{-2}$)

• A. $10m{s}^{-1}$
• B. $14m{s}^{-1}$
• C. $20m{s}^{-1}$
• D. $28m{s}^{-1}$

Answer 1

Answer: (C)

$20m{s}^{-1}$

Initial energy $E_i=PE+KE=mgh+\frac{1}{2}m{v}^2$
Energy when it reaches the ground: $E=\frac{1}{2}m{v}^{\prime 2}$ since PE is zero.
Given, $E=\frac{1}{2}{E}_i$
Since it rises to the same height $E_f=mgh+0$ is the final energy
$\Rightarrow \frac{1}{2}(mgh+\frac{1}{2}m{v}^2)=E_f=mgh+0$
$v^2=2gh=2\times 10\times 20={20}^2$
$\Rightarrow v=20m/s$