A ball falls from a height of $1\text{m}$ on the ground and jumps upto a height $H$. If the coefficient of restitution of the collision is $0.9$, then find the value of $H$.
(Take $g=10m/s^2$)

The correct option is B $0.81\text{m}$
Let,
$u_1=$ Speed of ball before collision with the surface
$v_1=$ Speed of ball after collision with surface
$u_2=$ Speed of ground before collision $=0$
$v_2=$ Speed of ground after collision $=0$
The coefficient of restitution, $e=\frac{\text{Speed of separation}}{\text{Speed of approach}}$
Speed of separation $=v_2-(-v_1)=v_1$ .....(i)
Speed of approach $u_1-u_2=u_1$ ....(ii)
$\therefore e=\frac{v_1}{u_1}$ ....(iii)
Since ball is dropped from height of $1\text{m}$, its speed while just colliding with the ground is
$u_1=\sqrt{2gh}=\sqrt{2\times 10\times 1}=\sqrt{20}\text{m/s}$
From eq (iii), we get
$0.9=\frac{v_1}{\sqrt{20}}$
$\Rightarrow v_1=\frac{9\sqrt{5}}{5}\text{m/s}$

Now, using $v^2-u^2=2aH$
Initial speed $u=v_1=\frac{9\sqrt{5}}{5}\text{m/s}$
Final speed $v=0,a=-g$
$0-{\left(\frac{9\sqrt{5}}{5}\right)}^2=2\times (-10)\times H$
$\Rightarrow H=0.81\text{m}$