Question

A ball is dropped on the floor from a height of $10\text{m}$. It rebounds to a height of $2.5\text{m}$. If the ball is in contact with the floor for $0.01\text{sec}$, the average acceleration during contact is (Neglect air friction and take $g=9.8m/s^2$)

• A. $2100{\text{m/s}}^2$ downwards
• B. $2100{\text{m/s}}^2$ upwards
• C. $1400{\text{m/s}}^2$
• D. $700{\text{m/s}}^2$

Answer 1

The correct option is B $2100{\text{m/s}}^2$ upwards

If vertical upwards direction is taken as positive y axis, then
Velocity at the time of striking the floor
$v_1=\sqrt{2g{h}_1}$
$=\sqrt{2\times 9.8\times 10}=14\text{m/s}$

Velocity with which it rebounds
$v_2=\sqrt{2g{h}_2}$
$=\sqrt{2\times 9.8\times 2.5}=7\text{m/s}$

Change in velocity $△v=v_2-v_1$
$=7-(-14)=21\text{m/s}$
(taking upward direction as +ve)

$\therefore$ Acceleration $=\frac{△v}{△t}$
$=\frac{21}{0.01}=2100{\text{m/s}}^2$ upwards