Question

A particle is projected in an x-y plane with a 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.

Answer 1

Hint: When a particle is projected in the x-y plane with the y-axis along vertical the particle travels in a parabolic path with the axis of the parabola parallel to the y-axis. When the y coordinate, the x coordinate, the angle of projection, and the speed of projection are connected through a mathematical equation this equation is known as the equation of that projectile motion.
Formula used:
$y = x\tan \theta - \dfrac{{g{x^2}}}{{2{u^2}{{\cos }^2}\theta }}$
where $y$ is the y coordinate, $x$ is the x coordinate, $\theta $ is the angle of projection, and $u$ is the speed of projection.

Complete step-by-step answer:
We know that when the y-coordinate, the x- coordinate, the angle of projection, and the speed of projection are connected through a mathematical equation this equation is known as the equation of that projectile motion and this is represented as
$y = x\tan \theta - \dfrac{{g{x^2}}}{{2{u^2}{{\cos }^2}\theta }}$
Where $y$ is the y coordinate, $x$ is the x coordinate, $\theta $ is the angle of projection, and $u$ is the speed of projection.
The equation of projectile is given in the question. Hence on comparing both the equation we can say that,
$\tan \theta = \sqrt 3 $
And
 ${u^2}{\cos ^2}\theta = 1$
We know that
$\tan (60^\circ ) = \sqrt 3 $
Hence on comparing we can say that the angle of projection is $60^\circ $.
Therefore
${u^2}{\cos ^2}(60^\circ ) = 1$
$ \Rightarrow {u^2} \times {(\dfrac{1}{2})^2} = 1$
As $\cos (60^\circ ) = \dfrac{1}{2}$
$\therefore {u^2} \times \dfrac{1}{4} = 1$
$ \Rightarrow {u^2} = 4$
$ \Rightarrow u = 2$
In the question, units are not specified so considering that all are in S.I. unit $u = 2m{s^{ - 1}}$

Therefore the speed of projection is $2m{s^{ - 1}}$.

Note:
The equation of projectile motion we used as the formula is only applicable if the point of projection is considered as the origin. Also, this formula can be easily proved by using Newton’s equation of motion on the motion of a particle along the y-axis and x-axis.