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The speed of a projectile at its maximum height is $\frac{\sqrt{3}}{2}$ times its initial speed. If the range of the projectile is p times the maximum height attained by it, then p =

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4/3

3/4

4 \sqrt{3}

2 \sqrt{3}

answer is C.

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Detailed Solution

V at maximum height

= \frac{\sqrt{3}}{2} u

$u \cos θ = \frac{\sqrt{3}}{2} u$

$\cos θ = \frac{\sqrt{3}}{2}$

$θ = 30 °$

$R = {PH}_{max}$

$\Rightarrow \sin 60 ° = P \frac{\sin^{2} 30 °}{2}$

$(\frac{\sqrt{3}}{2}) = \frac{P {(\frac{1}{2})}^{2}}{2}$

$\frac{\sqrt{3}}{2} = \frac{P}{2} (\frac{1}{4})$

$P = 4 \sqrt{3}$

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