The escape velocity from earth is 11.2 km per sec. If a body is to be projected in a direction making an angle `45^(@)` to the vertical, then the escape velocity is

Options

A. `11.2 xx 2` km/sec
B. `11.2` km/sec
C. `11.2 xx (1)/(sqrt(2))` km/sec
D. `11.2 xx sqrt(2)` km/sec

Explanation

To solve the problem, we need to determine the escape velocity when a body is projected at an angle of 45 degrees to the vertical. The escape velocity from the Earth is given as 11.2 km/s.

### Step-by-Step Solution:

1. **Understand Escape Velocity**:
The escape velocity from a celestial body is the minimum velocity needed for an object to break free from the gravitational attraction of that body without any further propulsion. For Earth, this value is given as 11.2 km/s.

2. **Escape Velocity Formula**:
The escape velocity (v_e) can be derived from the formula:
\[ v_e = \sqrt{\frac{2GM}{R}} \]
where G is the universal gravitational constant, M is the mass of the Earth, and R is the radius of the Earth. This formula shows that escape velocity depends only on the mass and radius of the Earth.

3. **Angle of Projection**:
The question specifies that the body is projected at an angle of 45 degrees to the vertical. However, it is important to note that escape velocity is independent of the angle of projection. This means that regardless of the angle at which the body is projected, 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 45 degrees to the vertical will still be 11.2 km/s.

Thus, the escape velocity when projected at an angle of 45 degrees to the vertical is **11.2 km/s**.

### Final Answer:
The escape velocity is **11.2 km/s**.

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