The escape velocity of a body of mass 1kg when projected from the surface of the earth vertically is `11.2"kms"^(-1)`. The escape velocity of a body of 1gm when projected at an angle `30^@` with the horizontal is

Options

A. `11.2/sqrt2 "kms"^(-1)`
B. `11.2 "kms"^(-1)`
C. `11.2 "kms"^(-1)`
D. `11.2/2` m/s

Answer

To solve the problem, we need to determine the escape velocity of a body of mass 1 gram when projected at an angle of 30 degrees with the horizontal.

### Step-by-Step Solution:

1. **Understand the Concept of Escape Velocity**:
The escape velocity is the minimum velocity required for an object to break free from the gravitational attraction of a celestial body without any further propulsion. The formula for escape velocity (V_e) is given by:
\[ V_e = \sqrt{2gR} \]
where \( g \) is the acceleration due to gravity and \( R \) is the radius of the celestial body.

2. **Given Data**:
- The escape velocity for a body of mass 1 kg projected vertically from the surface of the Earth is given as \( 11.2 \, \text{km/s} \).
- We need to find the escape velocity for a body of mass 1 gram projected at an angle of \( 30^\circ \).

3. **Mass Independence of Escape Velocity**:
It is important to note that escape velocity does not depend on the mass of the object. Therefore, whether the mass is 1 kg or 1 gram, the escape velocity remains the same.

4. **Angle Independence of Escape Velocity**:
Additionally, escape velocity is also independent of the angle of projection. This means that the escape velocity will be the same regardless of whether the object is projected vertically or at an angle (like \( 30^\circ \)).

5. **Conclusion**:
Since the escape velocity is independent of both mass and angle of projection, the escape velocity for a body of mass 1 gram projected at an angle of \( 30^\circ \) will still be:
\[ V_e = 11.2 \, \text{km/s} \]

### Final Answer:
The escape velocity of a body of 1 gram when projected at an angle of \( 30^\circ \) with the horizontal is \( 11.2 \, \text{km/s} \).

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