Question

The equation of a projectile is $y =\sqrt{3}x-\frac{gx^{2}}{2}$ The angle of projection is given by

• A. $\frac{\pi}{6}$
• B. $\frac{\pi}{3}$
• C. $\frac{\pi}{2}$
• D. zero

Answer 1

Answer: (B)

Here, the equation of projectile is
$y=\sqrt{3}x-\frac{gx^{2}}{2}$
Comparing the given equation with
$y=x\,tan\,\theta-\frac{gx^{2}}{2\upsilon^{2}\,cos^{2}\,\theta}$
we get, $tan \,\theta =\sqrt{3}$
$\Rightarrow \theta=tan^{-1}\left(\sqrt{3}\right)=\frac{\pi}{3}$