The path of a projectile is a parabola because the motion can be described by two independent components: horizontal motion at constant velocity and vertical motion under constant acceleration due to gravity. To show this mathematically, consider a projectile launched from the origin (0,0) with an initial velocity v0v_0v0​ at an angle θ\theta θ to the horizontal. The horizontal and vertical components of velocity are:

v0x=v0cos⁡θ,v0y=v0sin⁡θv_{0x}=v_0\cos \theta,\quad v_{0y}=v_0\sin \theta v0x​=v0​cosθ,v0y​=v0​sinθ

The position at time ttt is:

x=v0xt=v0cos⁡θ tx=v_{0x}t=v_0\cos \theta ,tx=v0x​t=v0​cosθt

y=v0yt−12gt2=v0sin⁡θ t−12gt2y=v_{0y}t-\frac{1}{2}gt^2=v_0\sin \theta ,t-\frac{1}{2}gt^2y=v0y​t−21​gt2=v0​sinθt−21​gt2

Eliminate ttt by expressing it from the xxx-equation:

t=xv0cos⁡θt=\frac{x}{v_0\cos \theta}t=v0​cosθx​

Substitute into the yyy-equation:

y=v0sin⁡θxv0cos⁡θ−12g(xv0cos⁡θ)2y=v_0\sin \theta \frac{x}{v_0\cos \theta}-\frac{1}{2}g\left(\frac{x}{v_0\cos \theta}\right)^2y=v0​sinθv0​cosθx​−21​g(v0​cosθx​)2

Simplify:

y=xtan⁡θ−g2v02cos⁡2θx2y=x\tan \theta -\frac{g}{2v_0^2\cos^2\theta}x^2y=xtanθ−2v02​cos2θg​x2

This is the equation of a parabola y=Ax−Bx2y=Ax-Bx^2y=Ax−Bx2, where A=tan⁡θA=\tan \theta A=tanθ and B=g2v02cos⁡2θB=\frac{g}{2v_0^2\cos^2\theta}B=2v02​cos2θg​ are constants. Thus, the path traced by a projectile under gravity (ignoring air resistance) is parabolic.