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Keeping the velocity of projection constant, the angle of projection is increased from 0° to 90º. Then the maximum height of the projectile

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decreases up to 90°

increase up to 45° and decreases beyond it

decreases up to 45° and increases beyond it

goes on increasing up to 90°

answer is A.

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Detailed Solution

$H = \frac{u^{2} \sin^{2} θ}{2 g}$

Height increases as $\sin θ$ increases which is maximum at 90^o.

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