The escape velocity for a body projected vertically upwards from the surface of the earth is `11.2 km s^(-1)`. If the body is projected in a direction making an angle `45^@` with the vertical, the escape velocity will be

Options

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

Answer

To solve the problem of 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 (VE) is the minimum velocity required for an object to break free from the gravitational pull of a celestial body without any further propulsion.

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

3. **Given Value**: From the problem, we know that the escape velocity when projected vertically upwards is \( 11.2 \, \text{km/s} \).

4. **Independence from Angle**: It is important to note that escape velocity is independent of the direction of projection. This means that whether the body is projected vertically or at an angle (like 45 degrees), the escape velocity remains the same.

5. **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:
\[ V_E = 11.2 \, \text{km/s} \]

### Final Answer:
The escape velocity when projected at an angle of 45 degrees with the vertical is **11.2 km/s**.

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