A body is projected upwards with a velocity of `3 xx 11.2 "km s"^(-1)` from the surface of earth. What will be the velocity of the body when it escapes from the gravitational pull of earth ?

Options

A. `11.2 "km s"^(-1)`
B. `sqrt(2) xx 11.2 "km s"^(-1)`
C. `sqrt(3) xx 11.2 "km s"^(-1)`
D. `sqrt(8) xx 11.2 "km s"^(-1)`

Explanation

To solve the problem, we will use the principle of conservation of energy. The total mechanical energy (kinetic + potential) of the body when it is projected upwards should equal the total mechanical energy when it escapes the gravitational pull of the Earth.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Initial velocity of the body, \( v_i = 3 \times 11.2 \, \text{km/s} \)
- Escape velocity from the surface of the Earth, \( v_e = 11.2 \, \text{km/s} \)

2. **Calculate Initial Velocity:**
\[ v_i = 3 \times 11.2 \, \text{km/s} = 33.6 \, \text{km/s} \]

3. **Use Conservation of Energy:**
The total mechanical energy at the surface of the Earth (initial) is equal to the total mechanical energy at infinity (final).
\[ \text{K.E.}_{\text{initial}} + \text{P.E.}_{\text{initial}} = \text{K.E.}_{\text{final}} + \text{P.E.}_{\text{final}} \]

At the surface of the Earth:
- Kinetic Energy (K.E.) = \( \frac{1}{2} m v_i^2 \)
- Potential Energy (P.E.) = \( -\frac{GMm}{R} \)

At infinity:
- K.E. = \( \frac{1}{2} m v_{\infty}^2 \)
- P.E. = 0 (since gravitational pull is negligible)

4. **Set Up the Equation:**
\[ \frac{1}{2} m v_i^2 - \frac{GMm}{R} = \frac{1}{2} m v_{\infty}^2 \]

5. **Cancel Mass (m) from the Equation:**
\[ \frac{1}{2} v_i^2 - \frac{GM}{R} = \frac{1}{2} v_{\infty}^2 \]

6. **Substituting Escape Velocity:**
The escape velocity \( v_e \) is given by:
\[ v_e = \sqrt{\frac{2GM}{R}} \]
Therefore, \( \frac{GM}{R} = \frac{1}{2} v_e^2 \).

7. **Substituting Back into the Equation:**
\[ \frac{1}{2} v_i^2 - \frac{1}{2} v_e^2 = \frac{1}{2} v_{\infty}^2 \]

8. **Multiply Through by 2:**
\[ v_i^2 - v_e^2 = v_{\infty}^2 \]

9. **Solve for \( v_{\infty} \):**
\[ v_{\infty}^2 = v_i^2 - v_e^2 \]
\[ v_{\infty} = \sqrt{v_i^2 - v_e^2} \]

10. **Substituting Values:**
\[ v_i = 33.6 \, \text{km/s}, \quad v_e = 11.2 \, \text{km/s} \]
\[ v_{\infty} = \sqrt{(33.6)^2 - (11.2)^2} \]
\[ v_{\infty} = \sqrt{1128.96 - 125.44} = \sqrt{1003.52} \approx 31.7 \, \text{km/s} \]

### Final Answer:
The velocity of the body when it escapes from the gravitational pull of Earth is approximately \( 31.7 \, \text{km/s} \).