Question

A tennis ball dropped from rest onto a cement floor rebounds $\frac{8}{10}$ of the height through which it fell. Neglecting air resistance, determine the coefficient of restitution.

• A. 0.8
• B. 0.895
• C. 0.75
• D. 0.5

Answer 1

The correct option is B 0.895

Let the ball be dropped through a height 'h'. The velocity of the ball just before impact,

$v_1=\sqrt{2gh}$

Since the floor is immovable, $v_2\&v_2^{\prime }=0$

$v_1^{\prime }-v_2^{\prime }=-e(v_1-v_2)$

$v_1^{\prime }=-e{v}_1=-e\sqrt{2gh}$

i.e., the ball rebounds with a velocity $e\sqrt{2gh}$. Now consider the motion of the ball after impact. It starts moving up with an initial velocity.

${\dot{x}}_0=e\sqrt{2gh}$

After reaching a height $x=\frac{8}{10}$ h, its velocity equals zero.

i.e., $\dot{x}=0$

and acceleration $\ddot{x}=-g$

using the relation ${\dot{x}}^2={\dot{x}}^2+2\ddot{x}x$

$i.e.,0=e^{2}2gh-\frac{8}{10}2gh$

$e^2=\frac{8}{10}$

$\Rightarrow e=0.895$