Keeping the velocity of projection constant, the angle of projection is increased from 00 to 900. Then the horizontal range of the projectile

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goes on increasing up to 900

decreases up to 900

increases up to 450 and decreases afterwards

decreases up to 450 and increases afterwards

answer is C.

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Detailed Solution

R=u2sin2αg'R' is maximum at α = 450'R' is zero at α = 00 and α =900so option - (3) is correct

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