Question

For a projectile thrown with a speed $v$, the horizontal range is $\frac{\sqrt{3}{v}^2}{2g}$. The vertical range is $\frac{v^2}{8g}$. The angle which the projectile makes with the horizontal initially is

• A. ${15}^{\circ }$
• B. ${30}^{\circ }$
• C. ${45}^{\circ }$
• D. ${60}^{\circ }$

Answer 1

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

We know that $R=\frac{v^2\sin 2\theta }{g}$
According to question
$\frac{v^2\sin 2\theta }{g}=\frac{\sqrt{3}{v}^2}{2g}$
or $\sin 2\theta =\frac{\sqrt{3}}{2}$ or $2\theta ={60}^{\circ }$ or $\theta ={30}^{\circ }$
Let us cross – check with the help of data for vertical range.
$\text{Vertical range}=\frac{v^2{\sin }^2\theta }{2g}$
$\frac{v^2{\sin }^2\theta }{2g}=\frac{v^2}{8g}$ or ${\sin }^2\theta =\frac{1}{4}$
or $\sin \theta =\frac{1}{2}$ or $\theta ={30}^{\circ }$
Hence, the correct answer is option (b)