The escape velocity for a body projected vertically upwards from the surface of earth is `11kms^(-1)`. If the same body is projected at an angle of `45^@` with the vertical , the escape velocity will be

Options

A. `11sqrt2kms^(-1)`
B. `22kms^(-1)`
C. `11kms^(-1)`
D. `11//sqrt2kms^(-1)`

Answer

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

### Step 1: Understand Escape Velocity
The escape velocity (v_e) is the minimum velocity required 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 upwards.

### Step 2: Recall the Formula for Escape Velocity
The formula for escape velocity is given by:
\[ v_e = \sqrt{2gR} \]
where:
- \( g \) is the acceleration due to gravity,
- \( R \) is the radius of the Earth.

### Step 3: Recognize Independence from Direction
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 any angle, the escape velocity remains the same.

### Step 4: Apply the Concept to the Given Angle
In this case, the body is projected at an angle of 45 degrees with the vertical. However, since escape velocity does not depend on the angle of projection, the escape velocity remains the same.

### Step 5: State the Final Answer
Therefore, the escape velocity when projected at an angle of 45 degrees with the vertical is still:
\[ v_e = 11 \text{ km/s} \]

### Conclusion
The escape velocity for the body projected at an angle of 45 degrees with the vertical remains 11 km/s.

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