Question

A ball of mass $50\text{g}$ is dropped from a height of $20\text{m}$. A boy on the ground hits this ball vertically upward with a bat with an average force of $200\text{N}$ so that it attains a vertical height of $45\text{m}$. The time for which the ball remains in contact with this bat is
$[$Take $g=10{\text{m/s}}^2]$

• A. $0.0325\text{s}$
• B. $0.0025\text{s}$
• C. $0.0125\text{s}$
• D. $0.0225\text{s}$

Answer 1

The correct option is C $0.0125\text{s}$


Velocity $\left(v_1\right)$ of ball when it falls through distance $h=20\text{m}$,
$v_1=\sqrt{2gh}=\sqrt{400}=20\text{m/s}$
Velocity of the ball after being hit vertically upward,
$v^2=u^2+2as$
$\Rightarrow 0=v_2^2-2\times 10\times 45$
$\Rightarrow v_2=\sqrt{900}=30\text{m/s}$
From the impulse momentum theorem,
$\overrightarrow{J}={\overrightarrow{P}}_f-{\overrightarrow{P}}_i$
$\Rightarrow +200\times \Delta t=0.05[30-(-20\left)\right]$
$[$taking upward as $+$ and downward as $-]$
$\Rightarrow 200\Delta t=0.05\times 50$
$\Rightarrow \Delta t=0.0125\text{s}$