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:

Maths; Projectiles; KS5 Year 13; Equation of the trajectory of a 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.