JEE MAIN - Physics (2003 - No. 32)
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 $${45^ \circ }$$ with the vertical, the escape velocity will be
$$11\sqrt 2 \,\,km/s$$
$$22$$ $$km/s$$
$$11$$ $$km/s$$
$${{11} \over {\sqrt 2 }}km/s$$
Explanation
We know, Escape velocity, $${v_e} = \sqrt {2gR} $$
So the escape velocity is independent of the angle at which the body is projected, hence it will remain same as 11 km/s.
So the escape velocity is independent of the angle at which the body is projected, hence it will remain same as 11 km/s.
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