Question

The equation of a projectile is $y=\sqrt{3}x-\dfrac{g{{x}^{2}}}{2}$ , find the angle of projection. Also find the speed of projection. Where at time t = 0, x = 0 and y = 0 also $\dfrac{{{d}^{2}}x}{d{{t}^{2}}}=0$ & $\dfrac{{{d}^{2}}y}{d{{t}^{2}}}=-g$

Answer 1

In the question the equation of motion of the projectile is given to us. Also, the initial conditions are given to us that the projectile at time t = 0, is at origin and the acceleration along the x-axis is zero while along the y-axis is equal to ‘g’. Comparing the given equation of the projectile, will enable us to determine the angle of projection and speed of projection as well.
Formula used:
$x=(u\cos \theta )t$
$y=(u\sin \theta )t-\dfrac{1}{2}g{{t}^{2}}$

Complete answer:
Let us say we project a body with initial velocity of ‘u’ at an angle $\theta$ with respect to the horizontal. Hence the vertical component of velocity is equal to $u\sin \theta$ and the horizontal component is equal to $u\cos \theta$ let us say at some instant of time ‘t’ the particle is at a point (x, y). The acceleration of the body along the x direction is zero i.e. $\dfrac{{{d}^{2}}x}{d{{t}^{2}}}=0$ and hence the body moves with constant velocity along the x-axis. Since distance covered (x) at some instant of time is equal to the product of speed into time we get,
$x=(u\cos \theta )t.....(1)$
The acceleration of the body along the y-axis is finite i.e. $\dfrac{{{d}^{2}}y}{d{{t}^{2}}}=-g$ . Hence the position of the particle along y-axis at some instant time ‘t’ using Newton’s second kinematic equation we get,
$y=(u\sin \theta )t-\dfrac{1}{2}g{{t}^{2}}....(2)$
Substituting for time ‘t’ in equation 1 from 2 we get,
$\begin{aligned} & y=(u\sin \theta )t-\dfrac{1}{2}g{{t}^{2}} \\\ & \because t=\dfrac{x}{u\cos \theta } \\\ & \Rightarrow y=(u\sin \theta )\left( \dfrac{x}{u\cos \theta } \right)-\dfrac{1}{2}g{{\left( \dfrac{x}{u\cos \theta } \right)}^{2}} \\\ & \therefore y=x\tan \theta -\dfrac{g{{x}^{2}}}{2{{u}^{2}}{{\cos }^{2}}\theta }....(3) \\\ \end{aligned}$
In the question the equation of the projectile is given as $y=\sqrt{3}x-\dfrac{g{{x}^{2}}}{2}$ . Comparing this with equation 3 we get,
$\begin{aligned} & \tan \theta =\sqrt{3} \\\ & \therefore \theta ={{60}^{\circ }} \\\ \end{aligned}$
Similarly comparing the coefficient of $g{{x}^{2}}$ we get,
$\begin{aligned} & -\dfrac{1}{2{{u}^{2}}{{\cos }^{2}}\theta }=-\dfrac{1}{2} \\\ & \Rightarrow \dfrac{1}{{{u}^{2}}{{\cos }^{2}}{{60}^{\circ }}}=1 \\\ & \Rightarrow {{u}^{2}}=\dfrac{1}{{{\cos }^{2}}{{60}^{\circ }}}=\dfrac{1}{{{\left( \dfrac{1}{2} \right)}^{2}}}=4 \\\ & \therefore u=2units/sec \\\ \end{aligned}$
Therefore, the angle of projection is 60 degrees and the speed of projection is 2 units per second.

Note:
It is to be noted that the equation of motion is nothing but a parabola. The vertex of the parabola is at the maximum height of the projectile. The height can be obtained by solving the differential form of the above equation taking the condition $\dfrac{dy}{dx}=0$ .