Motion in 2 and 3 dimensions

Projectile Motion is Parabolic

Can you show that $$v_x=v_{x0},\quad x=v_{x0}t,\quad v_y=v_{y0}-gt, \quad y =v_{y0}t-\frac{1}{2}g t^2\tag{1}$$ describes a parabola?

We start by noting that the initial velocity can be decomposed as $$ \begin{align*} v_{x0}&=v_0\cos(\alpha_0)\,,\\ v_{y0}&=v_0\sin(\alpha_0)\,. \end{align*} $$

Substituting into eqs. (1) we get $$\begin{align*} v_{x}&=v_0\cos(\alpha_0)\,,\nonumber\\ x&=v_0\cos(\alpha_0)\,t\,,\tag{2}\\ v_{y}&=v_0\sin(\alpha_0)-gt\nonumber\,,\\ y&=v_0\sin(\alpha_0)\,t-\frac{1}{2}gt \tag{3}\,. \end{align*}$$ At any time the distance of the projectile from the origin is given by the magnitude of the position vector $\vec{r}$, i.e., $$|\vec r |=r=\sqrt{x^2+y^2}\,,$$ and the projectile's speed is the magnitude of its velocity vector, i.e., $$|\vec v | =v=\sqrt{v_x^2+v_y^2}\,.$$

The direction of the velocity at any instant is (w.r.t. to the positive x-axis) given by $\alpha$, i.e., $$\alpha=\tan^{-1}\left(\frac{v_y}{v_x}\right)\,.$$

The shape of the projectile’s path (its trajectory) is now obtained by eliminating $t$ from the equations (2,3) for the $x$ and $y$ coordinates. From $$\begin{align*} x&=v_0\cos(\alpha_0)\,t\,,\\ y&=v_0\sin(\alpha_0)\,t-\frac{1}{2}gt^2\,, \end{align*} $$ it follows that $$t=\frac{x}{v_0\cos(\alpha_0)}$$ and thus $$\begin{align*} y &=v_0\sin(\alpha_0)\frac{x}{v_0\cos(\alpha_0)}-\frac{1}{2} g\frac{x^2}{v_0^2\cos^2(\alpha_0)}\\ &=\tan(\alpha_0)\, x-\frac{g}{2v_0^2\cos^2(\alpha_0)}x^2\,. \end{align*}$$

This is in the form $$y=bx-cx^2\,,$$ i.e. the equation of a parabola.