Terminal velocity is the constant speed a falling object reaches once air resistance balances its weight. Drag the sliders below to change the mass, cross-sectional area and drag coefficient, and watch the drag force climb and the speed level off in real time.
Interactive Lab · Mechanics
Drop the object and watch drag grow until it cancels weight. Time is compressed to a few seconds; the curve is the exact v = vt·tanh(g·t / vt).

The terminal velocity simulator is a free interactive physics lab that runs in your browser — nothing to install and no sign-up. Change the mass, cross-sectional area and drag coefficient and watch drag rise until it balances weight.
| Control | Range | Step |
|---|---|---|
| Mass | 1 – 150 kg | 1 |
| Cross-sectional area | 0.05 – 2 m² | 0.05 |
| Drag coefficient Cd | 0.1 – 1.5 | 0.05 |
Nothing falling through air can speed up without limit, and this simulation shows exactly where the acceleration runs out. Released from rest, the mass picks up speed, and as it does the drag force climbs with the square of that speed: F = 1/2 · rho · C · A · v². Watch the Drag force readout climb until it matches the object's weight. The instant drag equals weight, the net force hits zero, acceleration stops, and Current speed settles onto the value shown as Terminal velocity. The object is still falling and still has weight here — gravity and drag are balanced, not gone.
Now push the sliders. Raise Mass and terminal velocity rises; widen Cross-sectional area or nudge up Drag coefficient C and it drops, because those sit under the root in v_t = sqrt(2 · m · g / (rho · C · A)). Air density and gravity stay fixed on the panel, so every change you see traces back to those three inputs. The square root tempers each one: quadruple the mass and terminal velocity only doubles. The area and drag-coefficient terms soften the same way, so a skydiver tucking from a spread belly-down posture into a head-first dive cuts both the frontal area and the drag coefficient their body presents, and speeds up sharply — a real, sizeable jump, though the square root keeps it gentler than the drop in drag driving it.
Match your slider settings against the terminal velocity calculator to confirm the number, then wander into more falling, spinning, and oscillating systems waiting in our library of interactive physics sims.
The panel pins the air at 1.225 kg/m³, sea level on a standard day, and that one constant is the biggest thing the lab hides from a skydiver. Density falls with height: at a typical 4000 m exit the standard atmosphere gives 0.819 kg/m³, and because terminal velocity goes as 1/sqrt(rho), the default body — 75 kg, 0.70 m², C = 0.70, reading 49.5 m/s here — settles at 60.5 m/s up there, 22% faster. Terminal velocity is not one number for a jump: it falls all the way down, so most of a freefall is a gentle deceleration.
The drag coefficient is a dial here and a function of the flow in reality. C depends on the Reynolds number, and for a smooth sphere it collapses from about 0.47 to near 0.1 once the boundary layer turns turbulent — the drag crisis, near Re = 3 × 105. The dimples on a golf ball exist to trip that transition early so the ball flies on the low-drag side of it. The slider's own floor of 0.10 is roughly where a sphere lands after the crisis and 0.47 where it sits before: one object, two terminal velocities, and nothing here to say which you are in.
The v² law is itself the high-speed, inertia-dominated form of drag. Make the object small or slow enough and air stops being something to shove aside and starts behaving like syrup: drag turns proportional to v, and Stokes' law takes over. A 10 µm fog droplet falls at about 3 mm/s, while the formula on this page would predict 0.46 m/s for it — 150 times too fast. Cloud droplets hang in the air for hours precisely because they live in the regime this lab does not model.
Finally, the falling object is rigid, holds one attitude and drops through still air. Real bodies tumble, flutter and spin, and area and drag coefficient are not independent dials: a skydiver going from belly-down to head-down cuts both at once, which is why that change is dramatic rather than the modest shift two separate sliders imply. A thermal or a gust moves the whole balance, because drag depends on the body's speed through the air, not over the ground.
Sizing a parachute. The default setting reads 49.5 m/s, about 178 km/h — the familiar belly-to-earth freefall speed. Landing safely means dragging that to walking pace, and the only terms available are area and drag coefficient. Push the sliders to their draggiest corner, 2.00 m² at C = 1.50, and the best this lab manages is 20.0 m/s, still fatal. A real ram-air canopy presents roughly 25 m² at a drag coefficient near 1.5, and the same formula gives 5.7 m/s. A parachute is not a brake; it is an area multiplier of more than ten.
Rain. Hand the same expression a 2 mm raindrop — 4.19 mg of water, 3.1 mm² of frontal area, C about 0.5 for a sphere — and it returns 6.5 m/s. Drops do not keep accelerating over the kilometre they fall; they reach that speed in the first few metres and hold it. That is why rainfall intensity depends on how many drops arrive rather than how high the cloud sat.
Reading a thunderstorm from the ground. A 2 cm hailstone is 3.8 g of ice with 3.1 cm² of frontal area, and the same formula gives 19.6 m/s, about 71 km/h. Meteorologists run that backwards: a stone only grows to 2 cm if the updraft holding it aloft beats its own terminal velocity, so hail that size on the ground implies an updraft near 20 m/s inside the storm. Hail diameter is a physics-based readout of a storm's vertical wind.
Terminal velocity is the steady speed a falling object reaches when the drag force pushing up exactly balances its weight. The net force is then zero, so it stops accelerating and falls at a constant speed.
On the object's mass, its cross-sectional area and its drag coefficient (plus the air density): v_t = sqrt(2·m·g/(ρ·C·A)). A heavier object reaches a higher terminal speed; a larger area or drag coefficient lowers it.
Yes. Because v_t is proportional to the square root of the mass, more mass means a higher terminal speed for the same shape. The square root softens it, though — quadrupling the mass only doubles the terminal velocity.
Because drag grows with the square of the speed. As the object accelerates, drag rises until it equals the weight; from that moment the forces balance, the acceleration drops to zero, and the speed levels off at terminal velocity. The object is not weightless — gravity and drag simply cancel.
Strictly, never: the exact solution v = v_t · tanh(g·t/v_t) only approaches it. The yardstick is the time constant v_t/g, which is 5.0 s at the default settings — 76% of terminal velocity after 5 s, 96% after 10 s, 99.5% after 15 s. The on-screen drop is time-compressed so you need not wait.