Question

A ball is released from certain height which losses 50% if its kinetic energy on striking the ground, it will attain a height again.

• A. $\frac{1}{4}$ of initial height
• B. $\frac{1}{2}$ of initial height
• C. $\frac{3}{4}$ of initial height
• D. None of the above

Answer 1

Answer: (B)

$\frac{1}{2}$ of initial height

Let the initial height of ball is $h$ .

When it is released from $h$ , its velocity on ground ,

$v^2=0^2+2gh$ (initial velocity $u=0$)

or $v^2=2gh$

Hence ,its kinetic energy at ground ,

$K_1=\left(1/2\right)m{v}^2=\left(1/2\right)m\left(2gh\right)$

Now , let the height of ball is $h^{\prime }$ , after striking .

When ball got a height $h^{\prime }$ , after striking , its velocity after strike ,

$0^2=u^2-2g{h}^{\prime }$ (final velocity $v=0$)

or $u^2=2g{h}^{\prime }$

Hence , kinetic energy

$K_2=\left(1/2\right)m{u}^2=\left(1/2\right)m\left(2g{h}^{\prime }\right)$

As given ,its kinetic energy becomes $50$% after striking ,

$K_2=K_1\times 50$%

or $\left(1/2\right)m\left(2g{h}^{\prime }\right)$ $=\left(1/2\right)m\left(2gh\right)\times \left(1/2\right)$

or $h^{\prime }=h/2$