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A ball is projected with a velocity, $10 m s^{- 1}$ at an angle of $60 º$ with the vertical direction. Its speed at the highest point of its trajectory will be

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$Z e r o$

$10 m s^{- 1}$

$5 \sqrt{3} m s^{- 1}$

$5 m s^{- 1}$

answer is B.

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Detailed Solution

With horizontal the angle is 30º speed at maximum height is 10 cos 30

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