Question

A crazy ball is dropped on to the floor from the hand of the children from the height of $2m$. It rebounds to the height of $1.5m$. If the ball was in contact with the floor for $0.02$ second, what was the average acceleration during contact?

• A. $544.5m{s}^{-2}$
• B. $600m{s}^{-2}$
• C. $589.5m{s}^{-2}$
• D. $400m{s}^{-2}$

Answer 1

Answer: (C)

Average Acceleration $=\frac{{\vec{v}}_2-{\vec{v}}_1}{\Delta t}=\frac{\text{ Change in velocity }}{\text{ Change in time }}$
When ball falls from height $h_1$ we get,
$v_1^2=u_1^2+2gh$
$v_1=\sqrt{2g{h}_1}\dots \dots (1)(u=0$ under free fall)
Similarly, when ball falls from height $h_2$ after striking the fiont
But $g=-g$ because after striking ball goes upward against the gravity.
$v_2^2=u_2^2+2(-g)h_2$
(After contact $v_2=0$ at rest highest point, $u_2=v_2)$
$0=v_2^2-2g{h}_2$
$v_2^2=2gh\Rightarrow v_2=\sqrt{2g{h}_2}\dots ..(2)$
Average acceleration $=\frac{v_2-v_1}{\Delta t}$
$=\frac{v_2+v_1}{\Delta t}\left(v_1=-ve\right)$
Average acceleration $=\frac{\sqrt{2g{h}_2}+\sqrt{2g{h}_1}}{\Delta t}$
$=\frac{\sqrt{2\times 10\times 2}+\sqrt{2\times 10\times 1.5}}{0.020}$
Averageacceleration $=\frac{\sqrt{40}+\sqrt{30}}{0.020}=\frac{6.32+5.47}{0.020}$
Average acceleration $=\frac{11.79\times 100}{2}=\frac{1179}{2}$
$a_{av}=589.5m/s^2$