A particle is projected in an x-y plane with y-axis along vertical, the point of projection is the origin. The equation of a path is $y = \sqrt 3 x - \dfrac{{g{x^2}}}{2}$. Find the angle of projection and speed of projection.

Hint: We will find the general equation of a projectile projected at an angle $\alpha $ with the initial velocity imparted as $u$. This is obtained to be equal to $y = x\tan \alpha - \dfrac{{g{x^2}}}{{2{u^2}{{\cos }^2}\alpha }}$, where $x$ is the horizontal position and $y$ is the vertical displacement. We will compare the equation given in the question with this equation which has been obtained to get the required values of angle and speed of projection.
Formula used: The laws of motion equations $S = ut + \dfrac{1}{2}g{t^2}$ and $v = u + at$.

Complete Step by step answer
Let us assume that the angle of projection with which the particle has been initially projected as $\alpha $ and the initial velocity imparted to the particle as $u$.
Then from the x-component of this velocity, we will get
$x = (u\cos \alpha )t$
and from the y-component of this velocity, we will get
$y = (u\sin \alpha )t + \dfrac{1}{2}( - g){t^2}$.
Substituting the value of
$t = \dfrac{x}{{u\cos \alpha }}$
 from the above equation, we get
$y = (u\sin \alpha )(\dfrac{x}{{u\cos \alpha }}) + \dfrac{1}{2}( - g){(\dfrac{x}{{u\cos \alpha }})^2}$.
Thus, we get the general equation as
$y = x\tan \alpha - \dfrac{{g{x^2}}}{{2{u^2}{{\cos }^2}\alpha }}$,
 with the general assumption that the positive direction of vertical motion is upwards and that the gravity is downwards with a negative sign.
Comparing this with the equation given in the question,
$y = \sqrt 3 x - \dfrac{{g{x^2}}}{2}$ we get,
$\tan \alpha = \sqrt 3 $ and $2{u^2}{\cos ^2}\alpha = 2$.
Thus simplifying $\tan \alpha = \sqrt 3 $ and obtaining the principal value of the angle in degrees, we get, $\tan \alpha = \sqrt 3 = \tan \,{60^ \circ }$,
$ \Rightarrow \alpha = \,{60^ \circ }$
Substituting this value of $\alpha = \,{60^ \circ }$ in the other equation, we get
${u^2}{\cos ^2}{60^ \circ } = 1$
$ \Rightarrow {u^2}(\dfrac{1}{4}) = 1$ or
$u = 2$.
Therefore we get the angle of projection as
$\alpha = \,{60^ \circ }$
 and the speed of projection as
$u = 2$.

Note The above obtained equation of $y = x\tan \alpha - \dfrac{{g{x^2}}}{{2{u^2}{{\cos }^2}\alpha }}$ should be remembered or can be derived by eliminating the parameter of time. We can see that in this obtained equation when we substitute $y = 0$, we get a quadratic in $x$, which is the required parabola, or the equation of the curve as $x$ increases.