Question

A ball of mass 0.1 kg, initially at rest, dropped from height of 1m. Ball hits the ground and bounces off the ground. Upon impact with the ground, the velocity reduces by 20%. The height (in m) to which the ball will rise is

Answer 1

Velocity with which ball hits the ground
$V_1=\sqrt{2gh}=\sqrt{2\times 9.81\times 1}$
$=4.4294\frac{m}{s}$
and $V_2=0.8\times 4.4294=3.54\frac{m}{s}$
$\Rightarrow Now\frac{1}{2}m{V}_2^2$ is kinetic energy of ball which will be decreasing as it goes upwards (because work is done against gravity) and gravitational potential energy increase
$\Rightarrow \frac{1}{2}m{V}_2^2=mg{h}^{\prime }$
$h^{\prime }=\frac{V_2^2}{2g}=\frac{(3.54{)}^2}{2\times 9.81}=0.638m$