Question

A tennis ball is released from rest at a height $h$ above the ground. At each bounce $50\%$ of its K.E. is last to its surroundings. What is the height reacted by the ball after its second bounce?

• A. $h/8$
• B. $h/4$
• C. $h/2$
• D. Zero

Answer 1

The correct option is B $h/4$

Velocity attained after falling a height $h$ will be $v=\sqrt[2]{22gh}$

first Bounce-

remaining $KE=\frac{m{v}_1^2}{2}=\frac{1}{2}\times \frac{m{v}^2}{2}$

so velocity in upward direction will be $v_1=\frac{v}{\sqrt[2]{22}}$

so after the time when it return from its top height it will have $same$ velocity as $v_1$

Now

for second bounce-

incoming KE is $\frac{m{v}_1^2}{2}$

and after the bounce the remaining KE will be$\frac{m{v}_2^2}{2}=\frac{1}{2}\times \frac{m{v}_1^2}{2}$

so velocity in upward direction will be $v_2=\frac{v_1}{\sqrt[2]{22}}=\frac{v}{2}$

so maximum height attained after this bounce will be $H=\frac{(v_2{)}^2}{2g}=\frac{(v/2{)}^2}{2g}=\frac{v^2}{8g}=\frac{2gh}{8g}=\frac{h}{4}$