I dedicate this post to my younger brother, who is now learning about this sort of thing.
Note: I will assume that the reader has basic knowledge of trigonometry and calculus.
The simplicity of the laws of classical mechanics is so stupefying that it is almost criminal to candidly believe that the vastness of the universe, the cosmos, the celestial bodies can be described by a pithy set of mathematical aphorisms—and rightfully so, because quantum mechanics later tore (Newtonian) classical mechanics apart with its uncertainty principle, relativism, and (frankly bewildering) chaos.
Nevertheless, let us assume that we live in a utopian world (for projectiles) devoid of air friction and other "impurities," like cosmic rays, the Coriolis effect, humidity, and the gravitational pull of supernovae parsecs away, to mention a few things.
That said, let us work in a two-dimensional, euclidean, vector space defined by two orthogonal unit vectors $\mathbf i$ (horizontal, to the right) and $\mathbf j$ (vertical, up). These are our axes. Here is an illustration:
Let $\mathbf r(t)$, $\mathbf v(t)$, and $\mathbf a(t)$ describe the position, velocity, and acceleration, respectively, of our projectile at time $t$. Moreover, for the sake of simplicity, let us assume that, at time $t=0$, our projectile is launched from the origin, $\mathbf r(0)=\mathbf0$, at a speed of $s$ and at an angle of $\theta$ from the horizontal. Therefore, our initial velocity, $\mathbf v(0)$, can be described by the following vector:
$$\mathbf v(0)=s\cos\theta\mathbf i+s\sin\theta\mathbf j.$$
Here is an illustration of the above if it is not immediately obvious:
Now, here is where the magic happens: let us work with Newton's second law:
$$\mathbf F=m\mathbf a.$$
We already (tacitly) assumed that gravity ($-mg\mathbf j$) is the only force acting on our projectile. Therefore, from Newton's second law (rearranged), we deduce that
$$\begin{align}\mathbf a&=\frac{\mathbf F}m\\&=\frac{-mg\mathbf j}m\\&=-g\mathbf j.\end{align}$$
This is a simple differential equation that we can solve by integrating twice:
$$\begin{align}\mathbf v&=\int\mathbf a=\int-g\mathbf j=-gt\mathbf j+\mathbf C,\\\mathbf r&=\int\mathbf v=\int-gt\mathbf j+\mathbf C=-\frac g2t^2\mathbf j+\mathbf Ct+\mathbf D.\end{align}$$
Now it all boils down to finding the vectors $\mathbf C$ and $\mathbf D$, which is fairly straightforward:
$$\begin{align}\mathbf v(0)&=-g(0)\mathbf j+\mathbf C=\mathbf C=s\cos\theta\mathbf i+s\sin\theta\mathbf j,\\\mathbf r(0)&=-\frac g2(0)^2\mathbf j+(s\cos\theta\mathbf i+s\sin\theta\mathbf j)(0)+\mathbf D=\mathbf D=\mathbf 0.\end{align}$$
Therefore, the motion of our simple projectile is described by the following vector function:
$$\mathbf r(t)=\left[\left(-\frac g2t+s\sin\theta\right)\mathbf j+s\cos\theta\mathbf i\right]t.$$
Choosing some values $g$, $s$, and $\theta$ and plotting $\mathbf r$ will yield a nice, downward parabola delineating the path of our projectile. For example, letting $g=9.8\text{ m}\cdot\text{s}^{-2}$ (standard acceleration due to earth's gravity), $s=29\text{ m}\cdot\text{s}$, or $~65\text{ mi}\cdot\text{h}$, and $\theta=45^\circ$ yields the following trajectory (in meters):
Choosing some values $g$, $s$, and $\theta$ and plotting $\mathbf r$ will yield a nice, downward parabola delineating the path of our projectile. For example, letting $g=9.8\text{ m}\cdot\text{s}^{-2}$ (standard acceleration due to earth's gravity), $s=29\text{ m}\cdot\text{s}$, or $~65\text{ mi}\cdot\text{h}$, and $\theta=45^\circ$ yields the following trajectory (in meters):
I might in a later post introduce drag and wind factors. Stay tuned.