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

Answer

To solve the problem, we will use the concept of conservation of energy. The total mechanical energy (kinetic energy + potential energy) at the start will be equal to the total mechanical energy at the maximum height reached by the body.

### Step-by-Step Solution:

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

2. **Calculate the Initial Velocity:**
The body is projected with a velocity equal to \( \frac{3}{4} \) of the escape velocity:
\[ v = \frac{3}{4} v_s = \frac{3}{4} \sqrt{\frac{2GM}{R}} \]

3. **Initial Kinetic Energy (KE) and Potential Energy (PE):**
The initial kinetic energy \( KE_i \) when the body is projected is:
\[ KE_i = \frac{1}{2} mv^2 = \frac{1}{2} m \left(\frac{3}{4} v_s\right)^2 = \frac{1}{2} m \left(\frac{9}{16} v_s^2\right) = \frac{9}{32} m v_s^2 \]
The initial potential energy \( PE_i \) at the surface of the Earth is:
\[ PE_i = -\frac{GMm}{R} \]

4. **Total Initial Energy (E_i):**
The total initial energy \( E_i \) is the sum of the initial kinetic and potential energies:
\[ E_i = KE_i + PE_i = \frac{9}{32} m v_s^2 - \frac{GMm}{R} \]

5. **Final Kinetic Energy and Potential Energy at Maximum Height:**
At the maximum height \( h \), the final kinetic energy \( KE_f \) is zero (the body momentarily stops), and the potential energy \( PE_f \) is:
\[ PE_f = -\frac{GMm}{R+h} \]

6. **Total Final Energy (E_f):**
The total final energy \( E_f \) is:
\[ E_f = KE_f + PE_f = 0 - \frac{GMm}{R+h} \]

7. **Applying Conservation of Energy:**
By conservation of energy, we have:
\[ E_i = E_f \]
Substituting the expressions for \( E_i \) and \( E_f \):
\[ \frac{9}{32} m v_s^2 - \frac{GMm}{R} = -\frac{GMm}{R+h} \]

8. **Substituting for \( v_s^2 \):**
From the escape velocity formula, we know:
\[ v_s^2 = \frac{2GM}{R} \]
Therefore:
\[ \frac{9}{32} m \cdot \frac{2GM}{R} - \frac{GMm}{R} = -\frac{GMm}{R+h} \]

9. **Simplifying the Equation:**
\[ \frac{9GM}{16R} - \frac{GM}{R} = -\frac{GM}{R+h} \]
\[ \frac{9GM - 16GM}{16R} = -\frac{GM}{R+h} \]
\[ -\frac{7GM}{16R} = -\frac{GM}{R+h} \]

10. **Cross-Multiplying:**
\[ 7(R + h) = 16R \]
\[ 7h = 16R - 7R \]
\[ 7h = 9R \]
\[ h = \frac{9R}{7} \]

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