Terminal Velocity: When Drag Finally Catches Gravity

Two forces, one of them constant, one growing

A falling object feels gravity pulling down at a fixed mg, and air resistance pushing back up, growing as it speeds up. In the regime everyday objects fall through, drag is well modelled as proportional to the square of speed:

F_drag = ½ · ρ · v² · Cd · A

ρ  = density of the fluid (air ≈ 1.225 kg/m³ at sea level)
v  = speed relative to the fluid
Cd = drag coefficient (shape-dependent: ≈0.47 sphere, ≈1.0-1.3 skydiver belly-down)
A  = frontal (cross-sectional) area facing the flow

Where the two forces meet

Because drag scales with v² while gravity is fixed, there is exactly one speed at which they balance and net force drops to zero: the terminal velocity. Setting F_drag = mg and solving for v:

½ρv_t²CdA = mg

v_t = √( 2mg / (ρ·Cd·A) )

Everything here is fixed by the object's properties — v_t doesn't depend on how the object started or how far it has already fallen, only on its mass-to-area ratio and shape. Once it reaches v_t, net force is zero, so by Newton's first law it simply continues at that constant speed until something else changes (like deploying a parachute, which suddenly increases A and demands a whole new, much lower v_t).

The differential equation, and its exact solution

Writing k = ½ρCdA/m for convenience, Newton's second law for the fall becomes a first-order nonlinear ODE that, unlike the pendulum, does have a clean closed-form solution:

dv/dt = g - k·v²          starting from rest, v(0) = 0

v(t) = v_t · tanh( g·t / v_t )      where v_t = √(g/k)

tanh starts at 0 with slope g (pure free fall, as expected when v is still small and drag negligible) and asymptotically flattens to 1 as t grows — meaning v smoothly approaches v_t and never technically exceeds it starting from rest. That's exactly the shape the simulation's velocity-time graph traces out.

Why a feather, a ball and a skydiver disagree so much

v_t depends on m/A — mass per unit frontal area — not on mass alone. A feather has almost no mass but a relatively huge area, so its m/A is tiny and it reaches a very low terminal velocity, drifting rather than falling. A dense compact ball has a much higher m/A and needs a much higher speed before drag can catch up. A skydiver belly-down maximises A relative to their mass (roughly 50 m/s, about 180 km/h) but reorients head-down to shrink A and can exceed 90 m/s — the same body, same mass, wildly different terminal velocities purely from changing shape.

The classic (wrong) intuition, and why Galileo's experiment mattered

Aristotelian physics held that heavier objects simply fall faster, full stop — a belief that survives folk intuition to this day, largely because it's true in air for shapes with similar drag coefficients but different densities. In a vacuum, with no drag at all, every object accelerates at exactly g regardless of mass, famously demonstrated by dropping a feather and a hammer together on the Moon. Terminal velocity is the whole reason air makes that Moon experiment come out differently on Earth: it isn't that gravity treats masses differently, it's that drag does.

Frequently asked questions

Try it live

Everything above runs in your browser — open Terminal Velocity and change the parameters while it is running. Nothing is installed, nothing is uploaded, the whole model lives in one tab.