During a projectile motion, if the maximum height equals the horizontal range, then the angle of projection with the horizontal is
A
tan−11
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B
tan−12
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C
tan−13
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D
tan−14
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Solution
The correct option is Dtan−14 We know that,
Range, R=u2sin2θg
Maximum height, Hmax=u2sin2θ2g
According to the question, R=Hmax ⇒2u2sinθcosθg=u2sin2θ2g ⇒2cosθ=sinθ2 ⇒4=tanθ →θ=tan−14
So, option D is correct.