A ball is projected from the ground at an angle of $45^\circ$ with the horizontal surface. It reaches a maximum height of $120\,\text{m}$ 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 $30^\circ$ with the horizontal surface. The maximum height it reaches after the bounce, in metres, is ______.

Step 1: Relate the initial maximum height to the projection speed.

The maximum height $H_1$ of a projectile launched with speed $u$ at an angle of $45^\circ$ is given by the formula:

$$H_1 = \frac{u^2 \sin^2 45^\circ}{2g}$$

Substitute $H_1 = 120\text{ m}$ to find $u^2$:

$$120 = \frac{u^2 \left(\frac{1}{\sqrt{2}}\right)^2}{2g}$$ $$u^2 = 480g$$

Step 2: Find the speed of the ball immediately after the bounce.

The ball loses half of its kinetic energy upon impact with the ground:

$$\frac{1}{2}mv'^2 = \frac{1}{2}\left(\frac{1}{2}mu^2\right)$$ $$v' = \frac{u}{\sqrt{2}}$$

Step 3: Find the vertical component of the velocity after the bounce.

The velocity after the bounce makes an angle of $30^\circ$ with the horizontal:

$$v'_y = v'\sin 30^\circ$$ $$v'_y = \frac{u}{\sqrt{2}} \cdot \frac{1}{2}$$ $$v'_y = \frac{u}{2\sqrt{2}}$$

Step 4: Calculate the maximum height reached after the bounce.

The new maximum height $H_2$ is determined by the vertical component of the velocity:

$$H_2 = \frac{v'^2_y}{2g}$$ $$H_2 = \frac{\left(\frac{u}{2\sqrt{2}}\right)^2}{2g}$$ $$H_2 = \frac{u^2}{16g}$$

Substitute $u^2 = 480g$:

$$H_2 = \frac{480g}{16g}$$ $$H_2 = 30\text{ m}$$

Answer: $30$