Question

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

• A. $544.5\, ms ^{-2}$
• B. $600 \,ms ^{-2}$
• C. $589.5\, ms ^{-2}$
• D. $400 \,ms ^{-2}$

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,
$v _{1}^{2}= u _{1}^{2}+2 gh$
$v_{1}=\sqrt{2 g h_{1}} \ldots \ldots(1)(u=0$ 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, $\left.u_{2}=v_{2}\right)$
$0=v_{2}^{2}-2 g h_{2}$
$v_{2}^{2}=2 g h \Rightarrow v_{2}=\sqrt{2 g h_{2}} \ldots . .(2)$
Average acceleration $=\frac{v_{2}-v_{1}}{\Delta t}$
$=\frac{v_{2}+v_{1}}{\Delta t} \left(v_{1}=-v e\right)$
Average acceleration $=\frac{\sqrt{2 g h_{2}}+\sqrt{2 g h_{1}}}{\Delta t}$
$=\frac{\sqrt{2 \times 10 \times 2}+\sqrt{2 \times 10 \times 1.5}}{0.020}$
Averageacceleration $=\frac{\sqrt{40}+\sqrt{30}}{0.020}=\frac{6.32+5.47}{0.020}$
Average acceleration $=\frac{11.79 \times 100}{2}=\frac{1179}{2}$
$a_{ av }=589.5 \,m / s ^{2}$