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

Correct option is 3increases up to 450 and decreases afterwards

Questions

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

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detailed solution

Correct option is C

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