Question

A ball falling freely from a height of $4.9m/s$ hits a horizontal surface. If $e=\frac{3}{4},$ then the ball will hit the surface second time after :

• A. 0.5 s
• B. 1.5 s
• C. 3.5 s
• D. 3.4 s

Answer 1

Answer: (B)

From equation of motion
$v^2=u^2+2gh$ where $v$ is final velocity,
$u$ is initial velocity and $h$ is height
Given $h=4.9m,g=9.8m{s}^{-2},u=0$
$\therefore v=\sqrt{2gh}$
$=\sqrt{2\times 9.8\times 4.9}=9.8m{s}^{-1}$
Coefficient of restitution (e) of an object is a unit fraction value representing the ratio of velocities before and after an impact.
$\therefore v^{\prime }=\frac{3}{4}\times 9.8$
Time taken from first bounce to second bounce is
$t=\frac{2v^{\prime }}{g}$
$=2\times \frac{3}{4}\times 9.8\times \frac{1}{9.8}$
$=1.5s$