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

Answer 1

Answer: (B)

Velocity at the time of striking the floor,
$u=\sqrt{2g{h}_1}=\sqrt{2\times 9.8\times 10}=14m{s}^{-1}$
Velocity with which it rebounds.
$v=\sqrt{2g{h}_2}=\sqrt{2\times 9.8\times 2.5}=7m{s}^{-1}$
$\therefore$ Change in velocity $\Delta v=7-(-14)=21m{s}^{-1}$
Acceleration $=\frac{\Delta v}{\Delta t}=\frac{21}{001}=2100m{s}^{-2}$ (upwards)