Question

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

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

Answer 1

Answer: (B)

Escape speed of a body from Earth's surface is given by: $v_{min}=\sqrt{2gR}$
This expression is obtained by conservation of energy and doesn't involve in which direction the body is thrown/projected.
So, irrespective of the angle of projection, escape speed of the body from Earth's surface remains constant i.e. $\approx 11$ km/s