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 statements given and determine their validity based on the principles of physics, particularly focusing on projectile motion and conservation of angular momentum.

### Step-by-Step Solution:

1. **Understanding the Initial Conditions**:
- A particle is projected from the surface of the Earth with an initial velocity \( v_1 = 0.5 \sqrt{g R_e} \).
- The angle of projection with the vertical is \( 30^\circ \).

2. **Breaking Down the Velocity Components**:
- The initial velocity can be broken down into horizontal and vertical components.
- The vertical component \( v_{1y} = v_1 \cos(30^\circ) = 0.5 \sqrt{g R_e} \cdot \frac{\sqrt{3}}{2} = 0.25 \sqrt{3g R_e} \).
- The horizontal component \( v_{1x} = v_1 \sin(30^\circ) = 0.5 \sqrt{g R_e} \cdot \frac{1}{2} = 0.25 \sqrt{g R_e} \).

3. **Velocity at the Highest Point**:
- At the highest point of the projectile's motion, the vertical component of the velocity becomes zero, and only the horizontal component remains.
- Therefore, the velocity at the highest point \( v_2 = v_{1x} = 0.25 \sqrt{g R_e} \).

4. **Evaluating Statement 1**:
- Statement 1 claims that the velocity at the highest point will be \( 0.25 \sqrt{g R_e} \).
- From our calculations, we found that this is indeed true since \( v_2 = 0.25 \sqrt{g R_e} \).

5. **Conservation of Angular Momentum**:
- The angular momentum of the particle-Earth system is conserved.
- The initial angular momentum \( L_i \) can be expressed as \( L_i = m v_1 \cdot R_e \cdot \sin(30^\circ) \).
- The final angular momentum \( L_f \) at the highest point is \( L_f = m v_2 \cdot (R_e + h) \), where \( h \) is the height reached.

6. **Setting Up the Angular Momentum Conservation Equation**:
- Setting \( L_i = L_f \):
\[ m \cdot 0.5 \sqrt{g R_e} \cdot R_e \cdot \frac{1}{2} = m \cdot 0.25 \sqrt{g R_e} \cdot (R_e + h) \]
- Simplifying this equation leads to:
\[ 0.25 \sqrt{g R_e} \cdot R_e = 0.25 \sqrt{g R_e} \cdot (R_e + h) \]
- This implies that the height \( h \) is zero, indicating that the particle does not rise above the Earth's surface.

7. **Evaluating Statement 2**:
- Statement 2 claims that the angular momentum of the particle-Earth system remains constant, which is true as angular momentum is conserved in a gravitational field.

### Conclusion:
- **Statement 1** is **true**: The velocity at the highest point is indeed \( 0.25 \sqrt{g R_e} \).
- **Statement 2** is **true**: The angular momentum of the particle-Earth system remains constant.

### Final Answer:
Both statements are true.