The speed of a projectile at its maximum height is $\sqrt{3} / 2$ times its initial speed. If the range of the projectile is P times the maximum height attained by it, P is equal to

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4/3

$2 \sqrt{3}$

$4 \sqrt{3}$

3/4

answer is C.

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Detailed Solution

$Given \frac{\sqrt{3} u}{2} = ucos θ = speed at maximum height \cos θ = \frac{\sqrt{3}}{2} or θ = 30 ° Given mat {PH}_{\max} = R We know H_{\max} = \frac{Rtanθ}{4} P = \frac{R}{H_{\max}} = \frac{4}{tanθ} = \frac{4}{\tan 30 °} = 4 \sqrt{3}$

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