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Equation of the trajectory of a projectile

Equation of the trajectory of a projectile

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Summary

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, sss, in terms of time, ttt, 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, xxx and yyy, are both in terms of ttt. It is possible to find the Cartesian equation relating xxx and yyy, and this is known as the equation of trajectory.​



The equation of trajectory

A particle is projected with initial velocity UUU at an angle α\alphaα 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 xxx and vertical displacement yyy.


Horizontal motion

The horizontal component of the initial velocity is:

​ux=Ucos⁡αu_x=U \cos \alphaux​=Ucosα


The horizontal displacement is therefore:

​s=vtx=Ucos⁡α×t\begin{aligned}s &= vt \\x &= U \cos \alpha \times t\end {aligned}sx​=vt=Ucosα×t​​​


Vertical motion

The vertical component of the initial velocity is:

​uy=Usin⁡αu_y=U \sin \alphauy​=Usinα


Use suvatsuvatsuvat to find the vertical displacement:

s=ut+12at2y=Usin⁡α×t+12×(−9.8)×t2\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}sy​=ut+21​at2=Usinα×t+21​×(−9.8)×t2​​​


Equation of trajectory

Rearrange the equation with xxx for ttt​:

​t=xUcos⁡αt=\cfrac{x}{U\cos \alpha}t=Ucosαx​​​


Substitute into the equation with yyy​ to obtain:


​y=Usin⁡α×xUcos⁡α−12 g(xUcos⁡α)2y=U\sin\alpha \times \dfrac{x}{U\cos \alpha}-\dfrac{1}{2}\ g\Big (\dfrac{x}{U\cos \alpha}\Big)^2y=Usinα×Ucosαx​−21​ g(Ucosαx​)2​​


Since tan⁡α≡sin⁡αcos⁡α\tan\alpha\equiv\cfrac{\sin\alpha}{\cos\alpha}tanα≡cosαsinα​​ and 1cos⁡α≡sec⁡α\cfrac{1}{\cos\alpha}\equiv\sec\alphacosα1​≡secα​:


​y=xtan⁡α−gx22U2sec⁡2αy=x\tan\alpha-\cfrac{gx^2}{2U^2}\sec^2\alphay=xtanα−2U2gx2​sec2α​​


Finally, using sec⁡2α≡1+tan⁡2α\sec^2\alpha\equiv 1+\tan^2\alphasec2α≡1+tan2α substitute in for sec⁡2α\sec^2 \alphasec2α to obtain:


​y=xtan⁡α−gx2(1+tan⁡2α)2U2\boxed{y=x\tan\alpha -gx^2\cfrac{(1+\tan^2\alpha)}{2U^2}}y=xtanα−gx22U2(1+tan2α)​​​​


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.



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FAQs - Frequently Asked Questions

How do I work out the equation for the path of a projectile?

The equation for the path of a projectile can be derived from known formulae and manipulated using trigonometric identities.

What parameters affect the path of a projected particle?

The initial velocity, the angle of projection and the original point of projection all affect the path of a projected particle.

What is the equation of trajectory for a projectile?

The equation of trajectory is an equation showing the path of a projectile in terms of x (the horizontal displacement) and y (the vertical displacement).

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