Equation of the trajectory of a projectile

In a nutshell

As vertical and horizontal motion for a projectile is considered separately, equations for the displacement, $$s$$, in terms of time, $$t$$, can be obtained. These two equations, one for horizontal motion and one for vertical motion, can be considered to be parametric equations, as the horizontal and vertical displacements, $$x$$ and $$y$$, are both in terms of $$t$$. It is possible to find the Cartesian equation relating $$x$$ and $$y$$, and this is known as the equation of trajectory.​

The equation of trajectory

A particle is projected with initial velocity $$U$$ at an angle $$\alpha$$ above the horizontal. The diagram shows the path of the projectile:

Consider the particle on the path, having a horizontal displacement $$x$$ and vertical displacement $$y$$.

Horizontal motion

The horizontal component of the initial velocity is:

​$$u_x=U \cos \alpha$$

The horizontal displacement is therefore:

​$$\begin{aligned}s &= vt \\x &= U \cos \alpha \times t\end {aligned}$$​​

Vertical motion

The vertical component of the initial velocity is:

​$$u_y=U \sin \alpha$$

Use $$suvat$$ to find the vertical displacement:

$$\begin{aligned}s&= ut + \dfrac 1 2 a t^2 \\ \\y &= U \sin \alpha \times t + \dfrac 1 2 \times (-9.8) \times t^2\end {aligned}$$​​

Equation of trajectory

Rearrange the equation with $$x$$ for $$t$$​:

​$$t=\cfrac{x}{U\cos \alpha}$$​​

Substitute into the equation with $$y$$​ to obtain:

​$$y=U\sin\alpha \times \dfrac{x}{U\cos \alpha}-\dfrac{1}{2}\ g\Big (\dfrac{x}{U\cos \alpha}\Big)^2$$​​

Since $$\tan\alpha\equiv\cfrac{\sin\alpha}{\cos\alpha}$$​ and $$\cfrac{1}{\cos\alpha}\equiv\sec\alpha$$​:

​$$y=x\tan\alpha-\cfrac{gx^2}{2U^2}\sec^2\alpha$$​​

Finally, using $$\sec^2\alpha\equiv 1+\tan^2\alpha$$ substitute in for $$\sec^2 \alpha$$ to obtain:

​$$\boxed{y=x\tan\alpha -gx^2\cfrac{(1+\tan^2\alpha)}{2U^2}}$$​​

The equation of trajectory shows that the path of a projectile is dependent on the initial velocity and the angle of projection. The equation also confirms that the path of a projectile is in the shape of a parabola or quadratic curve.