The trajectory of a projectile is represented by y=(√3)x-(gx²)/2 . The angle of projection is

To find the angle of projection, we can compare the given trajectory equation of the projectile with the standard equation of a projectile's trajectory.

The trajectory equation of a projectile launched from the origin with an initial velocity $u$ at an angle of projection $\theta$ with the horizontal is given by:
$y=x\tan \theta -\frac{g{x}^2}{2u^2{\cos }^2\theta }$

The given equation of the trajectory is:
$y=\sqrt{3}x-\frac{g{x}^2}{2}$

By comparing the coefficients of $x$ in both equations, we get:
$\tan \theta =\sqrt{3}$

Since we know that $\tan (60°)=\sqrt{3}$, we have:
$\theta =60°$

Therefore, the angle of projection is 60°.