Question

The horizontal range of a projectile is $4\sqrt{3}$ times its maximum height. Its angle of projection will be

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

Answer 1

Answer: (D)

${30}^o$

Let Velocity is $V$ and angle of projection with horizontal is $\theta$.

$\therefore \text{Range}=\frac{2V^{2}sin\theta cos\theta }{g}$

$\therefore \text{Maximum Height}=\frac{V^{2}si{n}^2\theta }{2g}$

According to given condition, Range $=4\sqrt{3}$ Height

$\Rightarrow \frac{2V^{2}sin\theta cos\theta }{g}=4\sqrt{3}\frac{V^{2}si{n}^2\theta }{2g}$ $\Rightarrow tan\theta =\frac{1}{\sqrt{3}}$

$\Rightarrow \theta ={30}^o$