The escape velocity for a body projected vertically upwards from the surface of earth is $11 \mathrm{~km} /…
The escape velocity for a body projected vertically upwards from the surface of earth is $11 \mathrm{~km} / \mathrm{s}$. If the body is projected at an angle of $45^{\circ}$ with the vertical, the escape velocity will be
$11 \sqrt{2} \mathrm{~km} / \mathrm{s}$
$22 \mathrm{~km} / \mathrm{s}$
$11 \mathrm{~km} / \mathrm{s}$
$\frac{11}{\sqrt{2}} \mathrm{~km} / \mathrm{s}$
Solution
Escape velocity of a body is independent of the angle of projection. Hence, changing the angle of projection is not going to effect the magnitude of escape velocity