Maximum Height of Projectile After Inelastic Bounce at New Angle

To find the peak height reached by the ball after the bounce, we analyze the motion in three distinct steps:

Step 1: Motion before the first collision

Let $u_1$ be the initial speed of projection and $\theta_1 = 45^\circ$ be the angle of projection. The maximum height achieved during the initial trajectory is given by: $H_1 = \frac{u_1^2 \sin^2\theta_1}{2g}$

Given $H_1 = 120\text{ m}$ and $\theta_1 = 45^\circ$: $120 = \frac{u_1^2 \left(\frac{1}{\sqrt{2}}\right)^2}{2g} = \frac{u_1^2}{4g}$

Thus, we have: $\frac{u_1^2}{2g} = 240\text{ m}$

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Step 2: Energy loss during collision

Neglecting air resistance, the speed of the ball just before hitting the ground is equal to its initial speed $u_1$. Therefore, the kinetic energy right before impact is: $K_1 = \frac{1}{2} m u_1^2$

During the collision, the ball dissipates half of its total kinetic energy. The kinetic energy immediately after the bounce, $K_2$, is: $K_2 = \frac{1}{2} K_1$ $\frac{1}{2} m u_2^2 = \frac{1}{2} \left(\frac{1}{2} m u_1^2\right)$ $u_2^2 = \frac{1}{2} u_1^2$

where $u_2$ is the speed of the ball immediately after the bounce.

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Step 3: Motion after the bounce

After rebounding, the ball is projected at an angle $\theta_2 = 30^\circ$ relative to the horizontal. The new maximum height $H_2$ reached by the ball is: $H_2 = \frac{u_2^2 \sin^2\theta_2}{2g}$

Substituting $u_2^2 = \frac{1}{2} u_1^2$ and $\theta_2 = 30^\circ$: $H_2 = \frac{\left(\frac{1}{2} u_1^2\right) \sin^2(30^\circ)}{2g} = \frac{1}{2} \left(\frac{u_1^2}{2g}\right) \left(\frac{1}{2}\right)^2$

Using $\frac{u_1^2}{2g} = 240\text{ m}$: $H_2 = \frac{1}{2} \times 240 \times \frac{1}{4} = \frac{240}{8} = 30\text{ m}$

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Final Answer:

The peak height reached by the ball after the bounce is $30\text{ m}$ (or $30$).