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2nd Equation of Motion

The escape ve...

Question

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

A

\frac{11.2}{\sqrt{2}} k m s^{- 1}

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B

11.2 \times \sqrt{2} k m s^{- 1}

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C

11.2 \times 2 k m s^{- 1}

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D

11.2 k m s^{- 1}

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Solution

The correct option is B $11.2 k m s^{- 1}$
Change in energy $= 1 / 2 (G M m) / R) = 1 / 2 M v_{e}^{2}$
So it is independent of angle as gravitational field is conservative in nature.

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