A body is projected vertically upwards with a speed of ` sqrt((GM)/R)` ( M is mass and R is radius of earth ) . The body will attain a height of

Options

A. `R/2`
B. R
C. `5/4`R
D. `(3R)/2`

Answer

To solve the problem, we need to determine the maximum height attained by a body projected vertically upwards with an initial speed of \( u = \sqrt{\frac{GM}{R}} \), where \( M \) is the mass of the Earth and \( R \) is the radius of the Earth.

### Step-by-Step Solution:

1. **Identify the Initial Velocity**:
The initial velocity \( u \) of the body is given by:
\[ u = \sqrt{\frac{GM}{R}} \]

2. **Use the Conservation of Energy Principle**:
When the body is projected upwards, it will convert its kinetic energy into gravitational potential energy at the maximum height. The total mechanical energy at the start (kinetic energy) will equal the total mechanical energy at the maximum height (potential energy).

The kinetic energy (KE) at the start is:
\[ KE = \frac{1}{2} m u^2 \]

The potential energy (PE) at the maximum height \( h \) is:
\[ PE = -\frac{GMm}{R + h} \]

3. **Set Up the Energy Conservation Equation**:
At the maximum height, the total energy is conserved:
\[ \frac{1}{2} m u^2 = \frac{GMm}{R} - \frac{GMm}{R + h} \]

4. **Substitute the Value of \( u \)**:
Substitute \( u = \sqrt{\frac{GM}{R}} \) into the kinetic energy equation:
\[ \frac{1}{2} m \left(\frac{GM}{R}\right) = \frac{GMm}{R} - \frac{GMm}{R + h} \]

5. **Simplify the Equation**:
Cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ \frac{1}{2} \frac{GM}{R} = \frac{GM}{R} - \frac{GM}{R + h} \]
Rearranging gives:
\[ \frac{GM}{R + h} = \frac{GM}{R} - \frac{1}{2} \frac{GM}{R} \]
This simplifies to:
\[ \frac{GM}{R + h} = \frac{1}{2} \frac{GM}{R} \]

6. **Cross-Multiply to Solve for \( h \)**:
Cross-multiplying gives:
\[ 2GM = GM \frac{R + h}{R} \]
Simplifying leads to:
\[ 2R = R + h \]
Thus:
\[ h = R \]

7. **Final Result**:
The maximum height attained by the body is:
\[ h = R \]

### Conclusion:
The body will attain a height of \( R \).