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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 solution


Physics

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