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

Answer

To solve the problem step by step, we will analyze the motion of the ball before and after the bounce.

### Step 1: Determine the initial kinetic energy before the bounce.
The ball is projected at an angle of \(45^\circ\) and reaches a maximum height of \(120 \, \text{m}\). The maximum height \(H\) is given by the formula:
\[ H = \frac{u^2 \sin^2 \theta}{2g} \]
where:
- \(H = 120 \, \text{m}\)
- \(u\) is the initial velocity
- \(\theta = 45^\circ\)
- \(g = 9.81 \, \text{m/s}^2\) (acceleration due to gravity)

Rearranging the formula to find \(u\):
\[ u^2 = \frac{2gH}{\sin^2 \theta} \]
Substituting the values:
\[ u^2 = \frac{2 \cdot 9.81 \cdot 120}{\sin^2 45^\circ} \]
Since \(\sin 45^\circ = \frac{1}{\sqrt{2}}\):
\[ u^2 = \frac{2 \cdot 9.81 \cdot 120}{\left(\frac{1}{\sqrt{2}}\right)^2} = \frac{2 \cdot 9.81 \cdot 120 \cdot 2}{1} = 2 \cdot 9.81 \cdot 120 \cdot 2 \]
Calculating \(u^2\):
\[ u^2 = 2 \cdot 9.81 \cdot 240 = 4704.8 \]

### Step 2: Calculate the velocity just before the bounce.
The kinetic energy just before the bounce is:
\[ KE = \frac{1}{2} mv^2 \]
where \(v = u\). Thus, the kinetic energy before the bounce is:
\[ KE_{initial} = \frac{1}{2} m (u^2) = \frac{1}{2} m (4704.8) \]

### Step 3: Determine the kinetic energy after the bounce.
The ball loses half of its kinetic energy upon hitting the ground:
\[ KE_{final} = \frac{1}{2} KE_{initial} = \frac{1}{2} \left(\frac{1}{2} m (4704.8)\right) = \frac{1}{4} m (4704.8) \]

### Step 4: Find the new velocity after the bounce.
Using the kinetic energy formula:
\[ KE_{final} = \frac{1}{2} m v'^2 \]
Setting the two expressions for kinetic energy equal:
\[ \frac{1}{4} m (4704.8) = \frac{1}{2} m v'^2 \]
Canceling \(m\) and rearranging gives:
\[ v'^2 = \frac{1}{2} (4704.8) \cdot \frac{1}{4} = \frac{4704.8}{2} \cdot \frac{1}{4} = \frac{4704.8}{8} = 588.1 \]

### Step 5: Determine the maximum height after the bounce.
The maximum height after the bounce can be calculated using:
\[ H' = \frac{v'^2 \sin^2 \theta'}{2g} \]
where \(\theta' = 30^\circ\). Thus:
\[ H' = \frac{588.1 \cdot \sin^2 30^\circ}{2 \cdot 9.81} \]
Since \(\sin 30^\circ = \frac{1}{2}\):
\[ H' = \frac{588.1 \cdot \left(\frac{1}{2}\right)^2}{2 \cdot 9.81} = \frac{588.1 \cdot \frac{1}{4}}{19.62} \]
Calculating \(H'\):
\[ H' = \frac{147.025}{19.62} \approx 7.49 \, \text{m} \]

### Final Answer:
The maximum height the ball reaches after the bounce is approximately \(7.49 \, \text{m}\).