The path of a projectile is a parabola. This can be shown using the equations of motion under uniform acceleration (gravity). Consider a projectile launched with initial velocity uuu at an angle θ\theta θ with respect to the horizontal. The horizontal and vertical components of the initial velocity are:

ux=ucos⁡θ,uy=usin⁡θu_x=u\cos \theta,\quad u_y=u\sin \theta ux​=ucosθ,uy​=usinθ

The horizontal displacement at time ttt is:

x=uxt=ucos⁡θ⋅tx=u_xt=u\cos \theta \cdot tx=ux​t=ucosθ⋅t

The vertical displacement at time ttt is:

y=uyt−12gt2=usin⁡θ⋅t−12gt2y=u_yt-\frac{1}{2}gt^2=u\sin \theta \cdot t-\frac{1}{2}gt^2y=uy​t−21​gt2=usinθ⋅t−21​gt2

Solve the horizontal displacement equation for time ttt:

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

Substitute this into the vertical displacement equation:

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

Simplify:

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

This is an equation of the form:

y=Ax−Bx2y=Ax-Bx^2y=Ax−Bx2

where A=tan⁡θA=\tan \theta A=tanθ and B=g2u2cos⁡2θB=\frac{g}{2u^2\cos^2\theta}B=2u2cos2θg​ are constants. Since this is a quadratic equation in xxx, it represents a parabola. Hence, the path of the projectile is parabolic.

If more detailed derivations or examples are needed, they are also available in multiple sources.