STATEMENT-1 : A particle is projected with a velocity `0.5sqrt(gR_(e ))` from the surface of the earth at an angle of `30^(@)` with the vertical (at that point ) its velocity at the highest point will be `0.25sqrt(gR_(e ))`
because
STATEMENT-2 : Angular momentum of the particle earth system remains constant.

Options

A. Statement-1 is True, Statement-2 is True, Statement-2 is a correct explanation for Statement-1
B. Statement-1 is True, Statement-2 is True, Statement-2 is NOT a correct explanation for Statement-1
C. Statement-1 is True , Statement-2 is False
D. Statement-1 is False , Statement-2 is True

Answer

To solve the problem, we need to analyze the two statements provided:

**Statement 1**: A particle is projected with a velocity \(0.5\sqrt{gR_e}\) from the surface of the Earth at an angle of \(30^\circ\) with the vertical. Its velocity at the highest point will be \(0.25\sqrt{gR_e}\).

**Statement 2**: The angular momentum of the particle-Earth system remains constant.

### Step-by-step Solution:

1. **Identify the Initial Velocity Components**:
The initial velocity \(u\) of the particle is given as:
\[ u = 0.5\sqrt{gR_e} \]
The angle of projection with the vertical is \(30^\circ\). Therefore, we can find the horizontal and vertical components of the initial velocity:
- Horizontal component (\(u_x\)):
\[ u_x = u \sin(30^\circ) = 0.5\sqrt{gR_e} \cdot \sin(30^\circ) = 0.5\sqrt{gR_e} \cdot \frac{1}{2} = 0.25\sqrt{gR_e} \]
- Vertical component (\(u_y\)):
\[ u_y = u \cos(30^\circ) = 0.5\sqrt{gR_e} \cdot \cos(30^\circ) = 0.5\sqrt{gR_e} \cdot \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{4}\sqrt{gR_e} \]

2. **Determine the Maximum Height**:
At the highest point of the projectile's motion, the vertical component of the velocity becomes zero. The time taken to reach the highest point can be calculated using the vertical motion equations:
\[ v_y = u_y - g t \]
Setting \(v_y = 0\) at the highest point gives:
\[ 0 = \frac{\sqrt{3}}{4}\sqrt{gR_e} - g t \implies t = \frac{\sqrt{3}}{4g}\sqrt{gR_e} = \frac{\sqrt{3R_e}}{4} \]

3. **Calculate the Horizontal Distance**:
The horizontal component of the velocity remains constant throughout the motion. Therefore, at the highest point, the horizontal velocity is:
\[ v_x = u_x = 0.25\sqrt{gR_e} \]

4. **Final Velocity at the Highest Point**:
Since the vertical component of the velocity is zero at the highest point, the total velocity at that point is simply the horizontal component:
\[ v = v_x = 0.25\sqrt{gR_e} \]

5. **Conclusion for Statement 1**:
Thus, the velocity at the highest point is indeed \(0.25\sqrt{gR_e}\), confirming that Statement 1 is true.

6. **Conclusion for Statement 2**:
The angular momentum of the particle-Earth system remains constant due to the conservation of angular momentum in the absence of external torques. However, it does not provide a direct explanation for the velocity at the highest point. Therefore, while Statement 2 is true, it is not a correct explanation for Statement 1.

### Final Answer:
- **Statement 1 is true**.
- **Statement 2 is true but does not explain Statement 1**.