If the escape velocity of the body which is thrown from earth's surface is `V_e`, then the escape velocity of the body if it is thrown at an angle `theta` from horizontal earth`s surface is

Options

A. `V_e`
B. `V_ecostheta`
C. `V_esintheta`
D. `V_etantheta`

Answer

To solve the problem, we need to understand the concept of escape velocity and how it relates to the angle of projection.

### Step-by-Step Solution:

1. **Definition of 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 further propulsion. For Earth, this velocity is denoted as \( V_e \).

2. **Understanding the Angle of Projection**:
When a body is projected at an angle \( \theta \) with respect to the horizontal, we need to consider that the escape velocity is a scalar quantity. The direction of projection does not affect the magnitude of the escape velocity.

3. **Energy Consideration**:
The total mechanical energy of the body must be zero for it to escape the gravitational field. This means that the kinetic energy must equal the gravitational potential energy at the surface of the Earth. The escape velocity is derived from this energy balance.

4. **Independence from Angle**:
The escape velocity \( V_e \) is independent of the angle of projection. Whether the body is thrown vertically upwards or at an angle \( \theta \), the required speed to escape remains the same. The angle only affects the trajectory, not the speed needed to escape.

5. **Conclusion**:
Therefore, the escape velocity when thrown at an angle \( \theta \) from the horizontal remains \( V_e \).

### Final Answer:
The escape velocity of the body if it is thrown at an angle \( \theta \) from the horizontal Earth's surface is still \( V_e \).

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