Move a body through air and the air pushes back: $F_D = \tfrac{1}{2} C_d \rho A v^2$ — read aloud, F-D equals one-half C-d rho A v squared. Five symbols, five jobs. $F_D$ is the drag force in newtons. $C_d$ is the drag coefficient, a naked number set by shape alone — around 0.3 for a modern car, near 1.0 for a flat plate held broadside. $\rho$ (the Greek letter rho, said “roe”) is the fluid density in $\mathrm{kg/m^3}$; sea-level air is 1.225. $A$ is the frontal area in $\mathrm{m^2}$ — the silhouette the body presents to the flow, not its total surface. And $v$ is the speed through the fluid, in m/s.

The $v^2$ is the personality of this relation. Double the speed and the drag QUADRUPLES; the power needed to push through it goes up eightfold, because power is force times speed. That single fact explains highway fuel economy, why cyclists tuck, and why the last few knots of a ship's speed cost more than all the rest.

Now drop something. It speeds up, drag grows with the square of that speed, and very soon drag has grown to match the weight exactly. Net force zero, acceleration zero, speed constant forever after — terminal velocity. Set $\tfrac{1}{2} C_d \rho A v_t^2 = mg$ and solve: $v_t = \sqrt{\dfrac{2mg}{\rho A C_d}}$, where $v_t$ is that steady falling speed in m/s and $m$ is the falling mass in kilograms. Note what is NOT in it: the drop height. Past the point where terminal speed is reached, a longer fall changes nothing — which is the single most useful thing this relation will ever tell you. Note also the square root: doubling a parachute's area buys only $\sqrt{2}$ in landing speed, so chutes get big fast.