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The Equation for the Path of Projectile

The escape ve...

Question

The escape velocity for a body projected vertically upward from the surface of earth is $11 k m / s$ . If the body is projected at an angle of $45^{o}$ with the vertical, the escape velocity will be

A

11 \sqrt 2 k m / s

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B

22 k m / s

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C

11 k m / s

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D

11 \sqrt 2 m / s

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Solution

The correct option is C $11 k m / s$
Escape velocity does not depend upon the angle of projection So it is still $11 k m / s$ when thrown with angle of $45 °$

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The Equation for the Path of Projectile

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