Question

A ball falling freely from a height of $4.9 \,m/s$ 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
$v^{2}=u^{2}+2 g h$ where $v$ is final velocity,
$u$ is initial velocity and $h$ is height
Given $h=4.9\, m, g=9.8\, m s^{-2}, u=0$
$\therefore v=\sqrt{2 g h}$
$=\sqrt{2 \times 9.8 \times 4.9}=9.8\, ms ^{-1}$
Coefficient of restitution (e) of an object is a unit fraction value representing the ratio of velocities before and after an impact.
$\therefore v'=\frac{3}{4} \times 9.8$
Time taken from first bounce to second bounce is
$t=\frac{2 v'}{g}$
$=2 \times \frac{3}{4} \times 9.8 \times \frac{1}{9.8}$
$=1.5\, s$