Question

The escape velocity for a body projected vertically upwards from the surface of the earth is $11.2$ km $s^{-1}$. If the body is projected in a direction making an angle ${45}^o$ with the vertical, the escape velocity will be:

• A. $\frac{11.2}{\sqrt{2}}$km $s^{-1}$
• B. $11.2\times \sqrt{2}$km $s^{-1}$
• C. $11.2\times 2$km $s^{-1}$
• D. $11.2$km $s^{-1}$

Answer 1

The correct option is D $11.2$km $s^{-1}$

$V_e=\sqrt{2gr}$

It doesn't depend upon direction of projection.