The escape velocity for a body projected vertically upwards from the surface of earth is 11km/s. If the body is projected at an angle of `45^@` with the vertical, the escape velocity will be

Options

A. `11sqrt2km/s`
B. 22km/s
C. 11km/s
D. `(11)/(sqrt2) km/s`

Explanation

To solve the problem of finding the escape velocity when a body is projected at an angle of 45 degrees with the vertical, we can follow these steps:

### Step-by-Step Solution:

1. **Understand Escape Velocity**:
The escape velocity is the minimum velocity needed for an object to break free from the gravitational attraction of a celestial body without any further propulsion. For Earth, the escape velocity is given as 11 km/s when projected vertically.

2. **Formula for Escape Velocity**:
The formula for escape velocity (v_e) from the surface of a planet is given by:
\[ v_e = \sqrt{2gR} \]
where \( g \) is the acceleration due to gravity, and \( R \) is the radius of the Earth.

3. **Independence from Angle**:
It is important to note that escape velocity is independent of the angle of projection. This means that whether the body is projected vertically or at any angle, the escape velocity remains the same.

4. **Conclusion**:
Since the escape velocity does not depend on the angle of projection, the escape velocity when projected at an angle of 45 degrees with the vertical will still be:
\[ v_e = 11 \text{ km/s} \]

### Final Answer:
The escape velocity when the body is projected at an angle of 45 degrees with the vertical is **11 km/s**.

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