The escape velocity of a body projected vertically upward from the earth's surface is `11.2 " kms"^(-1)` . If the body is projected in a direction making `30^(@)` angle to the vertical, its escape velocity in this case will be

Options

A. `11.2 " kms"^(-1)`
B. `(11.2)/(2) " kms"^(-1)`
C. `11.2 xx (sqrt(3))/(2) "kms"^(-1)`
D. `(11.2)/(3)"kms"^(-1)`

Explanation

To solve the problem of finding the escape velocity of a body projected at an angle of \(30^\circ\) to the vertical, we can follow these steps:

### Step 1: Understand Escape Velocity
Escape velocity is the minimum velocity required for an object to break free from the gravitational attraction of a celestial body, without any additional propulsion. It is given by the formula:
\[ v_e = \sqrt{\frac{2GM}{R}} \]
where \(G\) is the gravitational constant, \(M\) is the mass of the Earth, and \(R\) is the radius of the Earth.

### Step 2: Given Escape Velocity
From the question, we know that the escape velocity when projected vertically upward from the Earth's surface is \(v_e = 11.2 \, \text{km/s}\).

### Step 3: Analyze the Angle of Projection
The problem states that the body is projected at an angle of \(30^\circ\) to the vertical. However, it is important to note that escape velocity is independent of the angle of projection. This means that whether the object is projected vertically or at any angle, the escape velocity remains the same.

### Step 4: Conclusion
Since the escape velocity does not depend on the angle of projection, the escape velocity when projected at \(30^\circ\) to the vertical will still be:
\[ v_e = 11.2 \, \text{km/s} \]

### Final Answer
The escape velocity of the body projected at an angle of \(30^\circ\) to the vertical is \(11.2 \, \text{km/s}\).

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