The escape speed of a body on the earth's surface is `11.2kms^(-1)`. A body is projected with thrice of this speed. The speed of the body when it escape the gravitational pull of earth is

Options

A. `11.2 kms^(-1)`
B. `22.4sqrt(2) kms^(-1)`
C. `(22.4)/(sqrt(2)) kms^(-1)`
D. `22.4sqrt(3) km s^(-1)`

Explanation

To solve the problem, we will use the principle of conservation of energy. The total mechanical energy of the body when it is projected must equal the total mechanical energy when it escapes the gravitational pull of the Earth.

### Step-by-Step Solution:

1. **Identify Given Values**:
- Escape speed of Earth, \( V_e = 11.2 \, \text{km/s} \).
- The body is projected with a speed \( V = 3V_e = 3 \times 11.2 \, \text{km/s} = 33.6 \, \text{km/s} \).

2. **Apply Conservation of Energy**:
The total energy at the surface when projected is equal to the total energy when it escapes:
\[ \frac{1}{2} mv^2 = \frac{1}{2} mv_0^2 + \frac{1}{2} mv_e^2 \]
Here, \( m \) is the mass of the body, \( v \) is the initial speed (33.6 km/s), \( v_0 \) is the speed when the body escapes, and \( v_e \) is the escape speed (11.2 km/s).

3. **Cancel the Mass**:
Since mass \( m \) is common in all terms, we can cancel it out:
\[ \frac{1}{2} v^2 = \frac{1}{2} v_0^2 + \frac{1}{2} v_e^2 \]
Simplifying gives:
\[ v^2 = v_0^2 + v_e^2 \]

4. **Rearrange to Find \( v_0 \)**:
We need to find \( v_0 \):
\[ v_0^2 = v^2 - v_e^2 \]
\[ v_0 = \sqrt{v^2 - v_e^2} \]

5. **Substitute the Values**:
Substitute \( v = 3V_e \) and \( v_e = 11.2 \, \text{km/s} \):
\[ v_0 = \sqrt{(3V_e)^2 - V_e^2} \]
\[ v_0 = \sqrt{(3 \times 11.2)^2 - (11.2)^2} \]
\[ v_0 = \sqrt{(33.6)^2 - (11.2)^2} \]

6. **Calculate the Squares**:
\[ (33.6)^2 = 1128.96 \quad \text{and} \quad (11.2)^2 = 125.44 \]
\[ v_0 = \sqrt{1128.96 - 125.44} \]
\[ v_0 = \sqrt{1003.52} \]

7. **Final Calculation**:
\[ v_0 \approx 31.7 \, \text{km/s} \]

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