Question

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

• A. $2100msec^{- 2}$ downwards
• B. $2100msec^{- 2}$ upwards
• C. $1400msec^{- 2}$
• D. $700msec^{- 2}$

Answer 1

Answer: (B)

Velocity at the time of striking the floor,
$u=\sqrt{2 g h_{1}}=\sqrt{2 \times 9 . 8 \times 10}=14ms^{- 1}$
Velocity with which it rebounds.
$v=\sqrt{2 g h_{2}}=\sqrt{2 \times 9 . 8 \times 2 . 5}=7ms^{- 1}$
$\therefore $ Change in velocity $\Delta v=7-\left(\right.-14\left.\right)=21ms^{- 1}$
Acceleration $=\frac{\Delta v}{\Delta t}=\frac{21}{001}=2100ms^{- 2}$ (upwards)