Question

A ball is dropped on the floor from a height of 10 m. It rebounds to a height of 2.5 m. If the ball is in contact with the floor for 0.01 sec, the average acceleration during contact is

• A.
• B.
• C.
• D.

Answer 1

The correct option is B

Velocity at the time of striking the floor,
$u=\sqrt{2g{h}_1}=\sqrt{2\times 9.8\times 10}=14m/s$

Velocity with which it rebounds.
$v=\sqrt{2g{h}_2}=\sqrt{2\times 9.8\times 2.5}=7m/s$

$\therefore$ Change in velocity $\Delta v=7-(-14)=21m/s$

Acceleration = $\frac{\Delta v}{\Delta t}=\frac{21}{0.01}=2100m/s^2$ (upwards)