A ball is projected from the  ground at an angle of ${45}^0$  with the horizontal surface .it reaches a maximum  height of  120m and  returns to the  ground  upon hitting the ground for the first time  it loses half  of its kinetic energy .immediately after the  bounce , the velocity of the ball makes an angle of ${30}^0$ with  the horizontal surface .the maximum height it reaches after the bounce ,in metres , is ----

$$\begin{gathered} H_0=\frac{{\left(v_0\sin {45}^0\right)}^2}{2g} \\ \Rightarrow \frac{v_0^2}{4g}=120m\mathrm{.....}\left(1\right) \end{gathered}$$

Upon hitting the ground for the first time it loses half of its kinetic energy.
Immediately  after the bounce the velocity of the ball makes an angle  of ${30}^0$  with  the horizontal surface 
Consider the maximum  height it reaches after the bounce=H and velocity of the projectile just after collision =v
So  $\frac{1}{2}m{v}^2=\frac{1}{2}\times \frac{1}{2}m{v}_0^2$
$\Rightarrow v=\frac{v_0}{\sqrt{2}}$
Maximum height reached by all after the bounce is ,
$$\begin{gathered} H=\frac{{\left(v\sin {30}^0\right)}^2}{2g} \\ H=\frac{v^2}{8g}=\frac{v_0^2}{16g} \\ \Rightarrow H=\frac{H_0}{4}=30m \end{gathered}$$