Question 110: Motion in 1D

Question

A tennis ball is dropped on to the floor from a height of 9.8 m. It rebounds to a height 5.0 m. Ball comes in contact with the floor for 0.2s. The average acceleration during contact is ___________ ms$$^{-2}$$.
(Given g = 10 ms$$^{-2}$$)

Accepted Answer

The speed of ball just before collision with ground is
$u=\sqrt{2 	imes g H}=\sqrt{2 	imes 10 	imes 9.8}=\underset{	ext { (Downwards) }}{14 \mathrm{~m} / \mathrm{sec}}$
The speed of ball just after collision is
$v=\sqrt{2 g h}=\sqrt{2 	imes 10 	imes 5}=\underset{	ext { (Upwards) }}{10 \mathrm{~m} / \mathrm{sec}}$
So, $\vec{a}=\frac{\Delta \vec{v}}{\Delta t}$
$=\frac{10+14}{0.2}=120 \mathrm{~m} / \mathrm{s}^{2}$

Motion in 1D

A tennis ball is dropped on to the floor from a height of 9.8 m. It rebounds to a height 5.0 m. Ball comes in contact with the floor for 0.2s. The average acceleration during contact is ___________ ms$^{-2}. (Given g = 10 ms^{-2}$)

Choose the correct answer: