A body is projected up with a velocity equal to `3//4th` of the escape velocity from the surface of the earth. The height it reaches is (Radius of the earth is `R`)

Options

A. `10R//9`
B. `9R//7`
C. `9R//8`
D. `10R//3`

Explanation

To solve the problem of finding the maximum height reached by a body projected upwards with a velocity equal to \( \frac{3}{4} \) of the escape velocity from the surface of the Earth, we can follow these steps:

### Step-by-Step Solution:

1. **Understand Escape Velocity**:
The escape velocity \( V_E \) from the surface of the Earth is given by the formula:
\[ V_E = \sqrt{\frac{2GM}{R}} \]
where \( G \) is the gravitational constant, \( M \) is the mass of the Earth, and \( R \) is the radius of the Earth.

2. **Calculate the Initial Velocity**:
The initial velocity \( v_0 \) of the body is:
\[ v_0 = \frac{3}{4} V_E = \frac{3}{4} \sqrt{\frac{2GM}{R}} \]

3. **Conservation of Energy**:
According to the conservation of mechanical energy, the total mechanical energy at the surface of the Earth (initial state) will be equal to the total mechanical energy at the maximum height \( h \) (final state).

- At the surface (initial):
- Kinetic Energy (KE) = \( \frac{1}{2} m v_0^2 \)
- Potential Energy (PE) = \( -\frac{GMm}{R} \)

- At maximum height \( h \) (final):
- Kinetic Energy (KE) = 0 (as the body comes to rest)
- Potential Energy (PE) = \( -\frac{GMm}{R+h} \)

4. **Set Up the Energy Conservation Equation**:
\[ \frac{1}{2} m v_0^2 - \frac{GMm}{R} = -\frac{GMm}{R+h} \]

5. **Substitute \( v_0 \)**:
Substitute \( v_0 = \frac{3}{4} \sqrt{\frac{2GM}{R}} \) into the equation:
\[ \frac{1}{2} m \left(\frac{3}{4} \sqrt{\frac{2GM}{R}}\right)^2 - \frac{GMm}{R} = -\frac{GMm}{R+h} \]
Simplifying the left side:
\[ \frac{1}{2} m \cdot \frac{9}{16} \cdot \frac{2GM}{R} - \frac{GMm}{R} = -\frac{GMm}{R+h} \]
\[ \frac{9GMm}{16R} - \frac{GMm}{R} = -\frac{GMm}{R+h} \]

6. **Combine Terms**:
Combine the terms on the left:
\[ \frac{9GMm}{16R} - \frac{16GMm}{16R} = -\frac{GMm}{R+h} \]
\[ -\frac{7GMm}{16R} = -\frac{GMm}{R+h} \]

7. **Cross-Multiply**:
Cross-multiplying gives:
\[ 7GMm(R+h) = 16GMmR \]
Dividing by \( GMm \) (assuming \( m \neq 0 \)):
\[ 7(R+h) = 16R \]

8. **Solve for \( h \)**:
Rearranging gives:
\[ 7h = 16R - 7R \]
\[ 7h = 9R \]
\[ h = \frac{9R}{7} \]

### Final Answer:
The maximum height \( h \) reached by the body is:
\[ h = \frac{9R}{7} \]