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show that the path of a projectile is a parabola

The path of a projectile (ignoring air resistance) is a parabola because its coordinates (x,y)(x,y)(x,y) satisfy an equation of the form y=ax+bx2y=ax+bx^{2}y=ax+bx2, which is the standard form of a parabola.

Set up the situation

Imagine you fire a projectile from the origin (0,0)(0,0)(0,0) with initial speed uuu at an angle θ\theta θ above the horizontal.

  • Let the horizontal axis be xxx, vertical axis be yyy.
  • Let ggg be the (constant) acceleration due to gravity acting downward.
  • Assume no air resistance, so there is no horizontal acceleration.

Initial velocity components:

  • Horizontal: ux=ucos⁡θu_{x}=u\cos \theta ux​=ucosθ.
  • Vertical: uy=usin⁡θu_{y}=u\sin \theta uy​=usinθ.

Horizontal and vertical motion

The motion splits cleanly into independent horizontal and vertical parts.

Horizontal motion (no horizontal acceleration):

x(t)=ucos⁡θ tx(t)=u\cos \theta ,tx(t)=ucosθt

Vertical motion (constant downward acceleration ggg):

y(t)=usin⁡θ t−12gt2y(t)=u\sin \theta ,t-\tfrac{1}{2}gt^{2}y(t)=usinθt−21​gt2

These two equations describe where the projectile is at any time ttt.

Eliminate time to get the trajectory

To find the path (i.e., a relation between yyy and xxx), remove ttt from the equations.

From the horizontal motion:

t=xucos⁡θt=\frac{x}{u\cos \theta}t=ucosθx​

Substitute this into the vertical equation:

y=usin⁡θ(xucos⁡θ)−12g(xucos⁡θ)2y=u\sin \theta \left(\frac{x}{u\cos \theta}\right) -\tfrac{1}{2}g\left(\frac{x}{u\cos \theta}\right)^{2}y=usinθ(ucosθx​)−21​g(ucosθx​)2

Simplify step by step:

  1. First term:

usin⁡θ⋅xucos⁡θ=x⋅sin⁡θcos⁡θ=xtan⁡θu\sin \theta \cdot \frac{x}{u\cos \theta} =x\cdot \frac{\sin \theta}{\cos \theta} =x\tan \theta usinθ⋅ucosθx​=x⋅cosθsinθ​=xtanθ

  1. Second term:

12g(xucos⁡θ)2=g2u2cos⁡2θx2\tfrac{1}{2}g\left(\frac{x}{u\cos \theta}\right)^{2} =\frac{g}{2u^{2}\cos^{2}\theta}x^{2}21​g(ucosθx​)2=2u2cos2θg​x2

So the trajectory equation becomes:

y=xtan⁡θ−g2u2cos⁡2θx2y=x\tan \theta -\frac{g}{2u^{2}\cos^{2}\theta}x^{2}y=xtanθ−2u2cos2θg​x2

Recognize the parabola

This equation has the form

y=ax+bx2y=ax+bx^{2}y=ax+bx2

where

  • a=tan⁡θa=\tan \theta a=tanθ
  • b=−g2u2cos⁡2θb=-\dfrac{g}{2u^{2}\cos^{2}\theta}b=−2u2cos2θg​

Both aaa and bbb are constants for a given launch (fixed uuu, θ\theta θ, and ggg). Therefore, yyy is a quadratic function of xxx, which is precisely the equation of a parabola.

So we have shown that the path of a projectile (under constant gravity and no air resistance) is a parabola.

Intuitive picture (short story)

Picture a hose spraying water at an angle on a calm day. Horizontally, each “droplet” moves equal distances every second because nothing speeds it up or slows it down in that direction. Vertically, though, gravity pulls it down harder and harder, so it drops more each second than it did the second before.

Combine “constant horizontal” with “accelerated vertical”, and the path bends into that familiar smooth arc—a parabola.

Information gathered from public forums or data available on the internet and portrayed here.