Question

A ball dropped from a height of $10$ m, rebounds to a height of $2.5$ m. If the ball is in contact with the floor for $0.01$ second, its acceleration during contact is ($g=9.8$ $m{s}^{-2}$)

• A. $20$ $m{s}^{-2}$
• B. $21$ $m{s}^{-2}$
• C. $210$ $m{s}^{-2}$
• D. $2100$ $m{s}^{-2}$

Answer 1

The correct option is D $2100$ $m{s}^{-2}$

When ball is hitting the ground,
It's velocity, $v$ is $=\sqrt{2gh}=\sqrt{2\times 9.8\times 10}$
After rebounding up its velocity be $u$.
$u^2=2gh\Rightarrow \sqrt{2\times g\times 2.5}$
So, $v=u+at$
$\Rightarrow \sqrt{2\times 9.8\times 10}=-\sqrt{2\times g\times 2.5}+a\left(0.01\right)$
$\Rightarrow \frac{\sqrt{2\times 9.8}}{0.01}(\sqrt{10}+\sqrt{2.5})=a$
$\Rightarrow a=2100m/s^2$