Q. 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 s, the average acceleration during contact is nearly [Take $g = 10ms^{- 2}$]
A) $500\sqrt{2}\ ms^{- 2}$upwards
B) 1800$\ ms^{- 2}$downwards
C) $1500\sqrt{5}\ ms^{- 2}$upwards
D) $1500\sqrt{2}\ ms^{- 2}$downwards
Explanation
(d) Average acceleration $= \frac{\mathrm{\Delta}v}{\mathrm{\Delta}t}$
$= \frac{\sqrt{2gh'} - \left( - \sqrt{2gh} \right)}{\mathrm{\Delta}t} = \frac{\sqrt{2gh'} + \sqrt{2gh}}{\mathrm{\Delta}t}$
$= \frac{\sqrt{2 \times 10 \times 2.5} + \sqrt{2 \times 10 \times 10}}{0.01}ms^{- 2}$
$= \frac{\sqrt{15} + \sqrt{200}}{0.01}\ ms^{- 2} = \frac{5\sqrt{2} + 10\sqrt{2}}{0.01}ms^{- 2}$
=$\frac{15\sqrt{2}}{0.01}ms^{- 2}$=1500$\sqrt{2}ms^{- 2}$
The upward velocity has been taken as positive. Since average acceleration is positive therefore its direction is vertically upward.
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.