A ball is projected form the ground at an angle of `45^(@)` with the horizonatl surface .It reaches a maximum height of 120 m and return to fthe 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^(@)` with the horizontal surface .The maximum height it reaches after the bounce in metres is _______

Explanation

To solve the problem step by step, we will follow the physics of projectile motion and energy conservation principles.

### Step 1: Determine the initial velocity (u) of the ball.
The maximum height (H) reached by the ball is given as 120 m. The formula for maximum height in projectile motion is:

\[ H = \frac{u^2 \sin^2 \theta}{2g} \]

Where:
- \( H = 120 \, \text{m} \)
- \( \theta = 45^\circ \)
- \( g = 10 \, \text{m/s}^2 \) (acceleration due to gravity)

Substituting the values:

\[ 120 = \frac{u^2 \left(\sin 45^\circ\right)^2}{2 \times 10} \]

Since \( \sin 45^\circ = \frac{1}{\sqrt{2}} \), we have:

\[ \sin^2 45^\circ = \frac{1}{2} \]

Thus, the equation becomes:

\[ 120 = \frac{u^2 \cdot \frac{1}{2}}{20} \]

Simplifying this gives:

\[ 120 = \frac{u^2}{40} \]

Multiplying both sides by 40:

\[ u^2 = 120 \times 40 = 4800 \]

### Step 2: Calculate the velocity after the bounce.
When the ball hits the ground, it loses half of its kinetic energy. The initial kinetic energy (KE_initial) is given by:

\[ KE_{\text{initial}} = \frac{1}{2} m u^2 \]

After losing half of its kinetic energy, the final kinetic energy (KE_final) is:

\[ KE_{\text{final}} = \frac{1}{2} KE_{\text{initial}} = \frac{1}{4} m u^2 \]

Thus, we can write:

\[ \frac{1}{2} m v^2 = \frac{1}{4} m u^2 \]

Canceling \( m \) from both sides and rearranging gives:

\[ v^2 = \frac{1}{2} u^2 \]

Substituting \( u^2 = 4800 \):

\[ v^2 = \frac{1}{2} \times 4800 = 2400 \]

### Step 3: Calculate the maximum height after the bounce.
After the bounce, the ball is projected at an angle of \( 30^\circ \). The formula for maximum height after the bounce is:

\[ H' = \frac{v^2 \sin^2 \theta'}{2g} \]

Where \( \theta' = 30^\circ \). Thus,

\[ H' = \frac{2400 \left(\sin 30^\circ\right)^2}{2 \times 10} \]

Since \( \sin 30^\circ = \frac{1}{2} \):

\[ \sin^2 30^\circ = \left(\frac{1}{2}\right)^2 = \frac{1}{4} \]

Substituting this back into the equation:

\[ H' = \frac{2400 \cdot \frac{1}{4}}{20} \]

This simplifies to:

\[ H' = \frac{600}{20} = 30 \, \text{m} \]

### Final Answer:
The maximum height the ball reaches after the bounce is **30 meters**.

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