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A projectile is thrown at a speed which is twice its speed at its maximum height. If R and H are its range and maximum height respectively then the ratio 2R/H is:

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$8 / \sqrt{3}$

$\sqrt{3} / 8$

$2 \sqrt{3}$

answer is A.

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Detailed Solution

$2 u \cos θ = u$

$\tan θ = \frac{1}{2}$

$\tan θ = \sqrt{3} / 1$

$R \tan θ = 4 H$

$\sqrt{3} H = 4 H$

$R / H = 4 / \sqrt{3}$

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