Question

the horizontal range of projectile is 2√3 times its Max Height find the angle of projection​

Answer 1

this is a direct formula question

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Answer 2

Explanation:

• u = initial velocity
• θ = angle of projectile

$\sf Height_{max} = \dfrac{u^2sin ^2 θ}{2g}$

According to the question,

$\sf \dfrac{2u^2 \sin θ \cos θ}{g}$

$\sf = 2 \sqrt{3} \bigg( \dfrac{u^2 \sin ^2 θ}{2g} \bigg)$

$\sf \implies \tan θ = \bigg( \dfrac{2}{\sqrt{3}} \bigg)$

$\sf θ =\tan ^{-1} \bigg( \dfrac{2}{\sqrt{3}} \bigg)$