1. Initial projection

The ball is projected at angle $45^\circ$ and reaches maximum height $120\,\text{m}$.

For a projectile, $H=\frac{u_y^2}{2g}$ where $u_y$ is the initial vertical component.

So, $120=\frac{u_y^2}{2g}$ $u_y^2=240g$

Since the launch angle is $45^\circ$, we have $u_x=u_y$ Hence just before hitting the ground for the first time:

• horizontal component remains $u_x$
• vertical component becomes $-u_y$

Therefore speed just before impact is $v^2=u_x^2+u_y^2=2u_y^2=2(240g)=480g$

2. Effect of bounce

On hitting the ground, the ball loses half of its kinetic energy. So its kinetic energy after bounce is half of that before bounce.

Since kinetic energy is proportional to $v^2$, $v'^2=\frac{1}{2}v^2=\frac{1}{2}(480g)=240g$

3. Velocity direction after bounce

Immediately after bounce, the velocity makes an angle $30^\circ$ with the horizontal.

Thus the vertical component after bounce is $v'_y=v'\sin 30^\circ$ So, $v_y'^2=v'^2\sin^2 30^\circ=240g\left(\frac{1}{2}\right)^2=240g\cdot \frac14=60g$

4. Maximum height after bounce

Now the ball rises with vertical component $v'_y$, so the maximum height reached after bounce is $H'=\frac{v_y'^2}{2g}=\frac{60g}{2g}=30$

Therefore, the maximum height after the bounce is $\boxed{30\,\text{m}}$