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Kepler's Law

The escape ve...

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 $45^{\circ}$ with the vertical, the escape velocity will be

A

11 \sqrt{2} k m s^{- 1}

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B

22 k m s^{- 1}

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C

11 k m s^{- 1}

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D

\frac{11}{\sqrt{2}} k m s^{- 1}

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Solution

The correct option is C $11 k m s^{- 1}$
Escape speed does not depend on the angle of projection

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