Question

For a projectile projected from ground at an angle $\theta$ with horizontal, $g{T}^2=\frac{2R}{\sqrt{3}}$, where $T$ is time of fight, $R$ is horizontal range of projectile, $g$ is acceleration due to gravity. The angle of projection $\left(\theta \right)$ is

• A. ${30}^{\circ }$
• B. ${60}^{\circ }$
• C. ${45}^{\circ }$
• D. ${90}^{\circ }$

Answer 1

The correct option is A ${30}^{\circ }$

According to question: $g{T}^2=\frac{2R}{\sqrt{3}}\Rightarrow g{\left(\frac{2u\sin \theta }{g}\right)}^2=\frac{\frac{2}{\sqrt{3}}{u}^2\sin 2\theta }{g}$
$g\left(\frac{4u^2{\sin }^2\theta }{g^2}\right)=\frac{2}{\sqrt{3}}\times \frac{u^2\sin 2\theta }{g}$
$\frac{4u^2{\sin }^2\theta }{g}=\frac{2}{\sqrt{3}}\times \frac{u^2\times 2\sin \theta \cos \theta }{g}$
On solving we get $\tan \theta =\frac{1}{\sqrt{3}}\Rightarrow \theta ={30}^{\circ }$