Question

The equation of a projectile is $y=\sqrt{3}x-\frac{g{x}^2}{2}$ 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
$y=\sqrt{3}x-\frac{g{x}^2}{2}$
Comparing the given equation with
$y=xtan\theta -\frac{g{x}^2}{2{\upsilon }^{2}co{s}^2\theta }$
we get, $tan\theta =\sqrt{3}$
$\Rightarrow \theta =ta{n}^{-1}\left(\sqrt{3}\right)=\frac{\pi }{3}$