A rocket of mass M is launched vertically from the surface of the earth with an initial speed `V= sqrt((gR)/(2))` . Assuming the radius of the earth to be R and negligible air resistance, the maximum height attained by the rocket above the surface of the earth is `(R)/(X)` where X is:

Explanation

To solve the problem of finding the maximum height attained by a rocket launched from the Earth's surface, we will use the principle of conservation of mechanical energy.

### Step-by-Step Solution:

1. **Identify the Given Values**:
- Initial speed of the rocket, \( V = \sqrt{\frac{gR}{2}} \)
- Radius of the Earth, \( R \)
- Mass of the rocket, \( M \) (not required for calculations)
- Gravitational acceleration at the surface of the Earth, \( g \)

2. **Write the Conservation of Energy Equation**:
The total mechanical energy at the launch point (initial) is equal to the total mechanical energy at the maximum height (final). Thus, we can write:
\[ U_i + K_i = U_f + K_f \]
Where:
- \( U_i \) = Initial potential energy
- \( K_i \) = Initial kinetic energy
- \( U_f \) = Final potential energy at maximum height
- \( K_f \) = Final kinetic energy at maximum height (which is 0 at maximum height)

3. **Calculate Initial Potential and Kinetic Energy**:
- Initial potential energy \( U_i = -\frac{GMm}{R} \)
- Initial kinetic energy \( K_i = \frac{1}{2} M V^2 = \frac{1}{2} M \left( \frac{gR}{2} \right) = \frac{MgR}{4} \)

4. **Calculate Final Potential Energy**:
At maximum height \( H \), the distance from the center of the Earth is \( R + H \):
\[ U_f = -\frac{GMm}{R + H} \]

5. **Set Up the Energy Conservation Equation**:
Substituting the energies into the conservation equation:
\[ -\frac{GMm}{R} + \frac{MgR}{4} = -\frac{GMm}{R + H} \]

6. **Simplify the Equation**:
Rearranging gives:
\[ -\frac{GMm}{R + H} = -\frac{GMm}{R} + \frac{MgR}{4} \]
Dividing through by \( GMm \) (assuming \( M \neq 0 \)):
\[ -\frac{1}{R + H} = -\frac{1}{R} + \frac{gR}{4GM} \]

7. **Substituting \( g \)**:
We know \( g = \frac{GM}{R^2} \), so substituting gives:
\[ -\frac{1}{R + H} = -\frac{1}{R} + \frac{R}{4R^2} = -\frac{1}{R} + \frac{1}{4R} \]
Simplifying further:
\[ -\frac{1}{R + H} = -\frac{4}{4R} + \frac{1}{4R} = -\frac{3}{4R} \]

8. **Cross-Multiplying**:
Cross-multiplying gives:
\[ 4R = 3(R + H) \]
Expanding and simplifying:
\[ 4R = 3R + 3H \implies R = 3H \implies H = \frac{R}{3} \]

9. **Expressing in Terms of X**:
The problem states that the maximum height \( H \) is given as \( \frac{R}{X} \). Thus:
\[ \frac{R}{3} = \frac{R}{X} \implies X = 3 \]

### Final Answer:
The value of \( X \) is \( 3 \).