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ABC is an equilateral triangle. From point A projectile was thrown such that it just grazes over B and reaches C. Given length of $A B = l$ . The angle of projection ‘ $θ$ ’ is $\tan^{- 1} (n \sqrt{3})$ , then n is ____

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answer is 2.

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Detailed Solution

$H = \frac{v^{2} \sin^{2} θ}{2 g}$

$\sin 60 = \frac{H}{l}$

$H = \frac{l \sqrt{3}}{2}$

$\frac{l \sqrt{3}}{2} = \frac{v^{2} \sin^{2} θ}{2 g}$

$U^{2} \sin^{2} θ = 10 \sqrt{3} l$

Range = $l = \frac{v^{2} s i n 2 θ}{g}$
$10 l = v^{2} \sin 2 θ$
$\frac{10 l}{\sin 2 θ} \sin^{2} θ = 10 \sqrt{3} l$

$θ = \tan^{- 1} (2 \sqrt{3})$

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