Q. A ball of mass $400\ gm$ is dropped from a height of $5m$. A boy on the ground hits the ball vertically upwards with a bat with an average force of $100\ newton$ so that it attains a vertical height of $20\ m$. The time for which the ball remains in contact with the bat is $\lbrack g = 10\ m/s^{2}\rbrack$
A) $0.12\ s$
B) $0.08\ s$
C) $0.04\ s$
D) $12\ s$
Explanation
(a) Velocity by which the ball hits the bat
$v_{1} = \sqrt{2gh_{1}} = \sqrt{2 \times 10 \times 5}$ or $\overrightarrow{v_{1}} = + 10m/s = 10\ m/s$
Velocity of rebound
$v_{2} = \sqrt{2gh_{2}} = \sqrt{2 \times 10 \times 20} = 20\ m/s$ or $\overrightarrow{v_{2}} = - 20\ m/s$
$F = m\frac{dv}{dt} = \frac{m(\overrightarrow{v_{1}} - \overrightarrow{v_{2}})}{dt} = \frac{0.4( - 10 - 20)}{dt} = 100\ N$
By solving $dt = 0.12\ \sec$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.