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Keeping the velocity of projection constant, the angle of projection is increased from 0º to 90º. Then the horizontal range of the projectile

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decreases up to 90º

goes on increasing up to 90º

increases up to 45º and decreases after-wards

decreases up to 45º and increases after-wards

answer is C.

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Detailed Solution

$R = \frac{u^{2} \sin (2 θ)}{g}$

$R_{\max} = \frac{u^{2}}{g} at θ = 45^{o}$

The horizontal range of the projectile increases up to 45º and decreases after-wards.

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