A bullet of mass `m` fired at `30^(@)` to the horizontal leaves the barrel of the gun with a velocity `upsilon`. The bullet hits a soft target at a height `h` above the ground while it is moving downward and emerges out with half the kinetic energy it had before hitting the target.
Which of the following statements are correct in respect of bullet after it emerges out of the target ?

Options

A. The velocity of the bullet remains the same.
B. The velocity of the bullet will be reduced to half its initial value
C. The velocity of the bullet will be more than half of its earlier velocity
D. The bullet will continue to move along the same parabolic path.

Explanation

To solve the problem step by step, we will analyze the bullet's motion and energy before and after it hits the target.

### Step 1: Understand the Initial Conditions
The bullet is fired at an angle of \(30^\circ\) with an initial velocity \(v\). The bullet will have both horizontal and vertical components of velocity.

**Hint:** Remember to break down the velocity into its horizontal and vertical components using trigonometric functions.

### Step 2: Calculate Initial Kinetic Energy
The initial kinetic energy (KE_initial) of the bullet can be calculated using the formula:
\[ KE_{\text{initial}} = \frac{1}{2} m v^2 \]
where \(m\) is the mass of the bullet and \(v\) is its initial velocity.

**Hint:** Kinetic energy is dependent on the square of the velocity, so any change in velocity will significantly affect the kinetic energy.

### Step 3: Determine Kinetic Energy After Hitting the Target
According to the problem, after hitting the target, the bullet's kinetic energy is reduced to half of its initial value:
\[ KE_{\text{final}} = \frac{1}{2} KE_{\text{initial}} = \frac{1}{4} m v^2 \]

**Hint:** This reduction in kinetic energy will also affect the velocity of the bullet after it emerges from the target.

### Step 4: Relate Kinetic Energy to Velocity After Hitting the Target
Using the kinetic energy formula for the bullet after it hits the target:
\[ KE_{\text{final}} = \frac{1}{2} m v_f^2 \]
Setting this equal to the reduced kinetic energy:
\[ \frac{1}{2} m v_f^2 = \frac{1}{4} m v^2 \]
We can cancel \(m\) from both sides (assuming \(m \neq 0\)):
\[ v_f^2 = \frac{1}{2} v^2 \]

**Hint:** When solving for \(v_f\), take the square root of both sides.

### Step 5: Solve for Final Velocity
Taking the square root gives:
\[ v_f = \frac{v}{\sqrt{2}} \]

**Hint:** This shows that the final velocity is not half of the initial velocity, but rather \(v\) divided by \(\sqrt{2}\).

### Step 6: Analyze the Options
1. **Velocity remains the same**: Incorrect, as the velocity changes.
2. **Velocity is reduced to half**: Incorrect, as we found \(v_f = \frac{v}{\sqrt{2}}\).
3. **Velocity is more than half of its initial value**: Correct, since \(\frac{v}{\sqrt{2}} > \frac{v}{2}\).
4. **Bullet continues along the same parabolic path**: Incorrect, as the bullet's velocity has changed, affecting its trajectory.

### Conclusion
The correct statement is that the bullet's velocity after it emerges from the target is more than half of its initial velocity.

**Final Answer:** The correct option is that the bullet's velocity after it emerges is more than half of its initial value.