Question

The escape velocity for a body projected vertically upwards from the surface of earth is $11km{s}^{-1}$. If the body is projected at an angle of ${45}^0$ with the vertical, the escape velocity will be:

• A. $11\sqrt{2}km{s}^{-1}$
• B. $22km{s}^{-1}$
• C. $11km{s}^{-1}$
• D. $11/\sqrt{2}m{s}^{-1}$

Answer 1

The correct option is C $11km{s}^{-1}$

The escape velocity is independent of angle of projections.

Hence changing the angle of projections doesn't affect the magnitude of escape velocity.

Hence it remains same as$11\frac{km}{s}$.