Q. A ball is dropped on the floor from a height of $10\ m$. 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) $2100\ m/\sec^{2}{downwards\ }$
B) $2100\ m/\sec^{2}{upwards\ }$
C) $1400\ m/\sec^{2}\ $
D) $700m/\sec^{2}$
Explanation
(b) Velocity at the time of striking the floor,
$u = \sqrt{2gh_{1}} = \sqrt{2 \times 9.8 \times 10} = 14m/s$
Velocity with which it rebounds
$v = \sqrt{2gh_{2}} = \sqrt{2 \times 9.8 \times 2.5} = 7\ m/s$
$\therefore Change\ in\ velocity\ \mathrm{\Delta}v = 7 - ( - 14) = 21m/s$
$\therefore Acceleration = \frac{\mathrm{\Delta}v}{\mathrm{\Delta}t} = \frac{21}{0.01} = 2100m/s^{2}(upwards)$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.