Question

# What is the angle $$\theta$$ of projection with horizontal plane of a projectile if its range is $$\frac { \sqrt { 3v^ 3 } }{ 2g }$$, where v is velocity of projection?

Solution

## Given that,Range $$R=\dfrac{\sqrt{3{{v}^{3}}}}{2g}$$ We know that,$$R=\dfrac{{{v}^{2}}\sin 2\theta }{g}$$ Now, the angle of projection   $$\dfrac{{{v}^{2}}\sin 2\theta }{g}=\dfrac{\sqrt{3{{v}^{3}}}}{2g}$$  $$\dfrac{{{v}^{4}}{{\sin }^{2}}2\theta }{{{g}^{2}}}=\dfrac{3{{v}^{3}}}{4{{g}^{2}}}$$  $${{\sin }^{2}}2\theta =\dfrac{3}{4v}$$  $$\sin 2\theta =\dfrac{\sqrt{3}}{2\sqrt{v}}$$  $$2\theta ={{\sin }^{-1}}\dfrac{\sqrt{3}}{2\sqrt{v}}$$  $$\theta ={{\sin }^{-1}}\dfrac{\sqrt{3}}{4\sqrt{v}}$$ Hence, this is the required solutionPhysics

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