The escape velocity for a body projected vertically upwards from the surface of the earth is $11.2{\rm{\;\;km}}{{\rm{s}}^{ - 1}}$. If the body is projected in a direction making an angle of ${45^0}$ with the vertical, the escape velocity will be

$11.2{\rm{\;\;km}}{{\rm{s}}^{ - 1}}$

$11.2 \times \sqrt 2 {\rm{\;km}}{{\rm{s}}^{ - 1}}$

$11.2 \times 2{\rm{\;km}}{{\rm{s}}^{ - 1}}$

$11.2/\sqrt 2 {\rm{\;km}}{{\rm{s}}^{ - 1}}$

JEE PHYSICS MCQ: GRAVITATION

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Understanding +4 | -1

The escape velocity for a body projected vertically upwards from the surface of the earth is

11.2\mathrm{\;\;km}{\mathrm{s}^{ - 1}}

. If the body is projected in a direction making an angle of

{45^0}

with the vertical, the escape velocity will be

View answer and explanation

Correct option: A

11.2\mathrm{\;\;km}{\mathrm{s}^{ - 1}}

Option A is correct because escape velocity is independent of the angle of projection.

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