Terminal Velocity Calculator (with air resistance)
Find the speed at which drag balances weight and a fall stops accelerating, from mass, drag coefficient, and frontal area.
Typical terminal velocities
Computed from the drag equation at sea-level air density (1.225 kg/m³). The masses and areas are representative rather than exact — change them above to match whatever you have in mind.
| Object | Cd | Area | Mass | Terminal velocity | km/h | mph |
|---|---|---|---|---|---|---|
| Skydiver, belly-down | 1 | 0.7 m² | 80 kg | 42.78 m/s | 154 | 95.69 |
| Skydiver, head-down dive | 0.7 | 0.18 m² | 80 kg | 100.8 m/s | 363 | 225.5 |
| Baseball (smooth sphere) | 0.47 | 0.004185386812745002 m² | 0.145 kg | 34.35 m/s | 123.7 | 76.85 |
| Cube, face-on | 1.05 | 0.01 m² | 0.5 kg | 27.61 m/s | 99.4 | 61.77 |
| Flat plate, face-on | 1.28 | 0.05 m² | 0.2 kg | 7.073 m/s | 25.46 | 15.82 |
| Streamlined body | 0.04 | 0.02 m² | 1 kg | 141.5 m/s | 509.3 | 316.5 |
Terminal velocity
A falling object does not keep speeding up indefinitely. Air resistance grows with the square of speed, so the faster it falls the harder the air pushes back — until the drag exactly cancels the weight and the acceleration reaches zero. From that point the speed is constant, and that speed is the terminal velocity.
vt = √( 2·m·g / (ρ·Cd·A) )
Mass and gravity push the speed up; air density, drag coefficient, and frontal area push it down. The square root matters: to fall twice as fast you need four times the mass, or a quarter of the area. This is why a parachute works at all — it barely changes the mass but multiplies the area enormously.
Worked example
An 80 kg skydiver falling belly-down presents roughly 0.7 m² with a drag coefficient near 1.0. That gives √(2 × 80 × 9.807 / (1.225 × 1.0 × 0.7)) ≈ 43 m/s, about 154 km/h or 96 mph — the usual quoted range for that posture. Turning head-down cuts the frontal area to something like 0.18 m² and the terminal velocity roughly doubles, which is exactly how speed skydivers go faster without getting heavier.
Mass matters here, unlike in free fall
The free fall calculator makes a point of mass cancelling out — a feather and a hammer fall identically in a vacuum. Add air and that stops being true. Drag depends on size and shape, weight depends on mass, and the balance between them is different for every object. The feather and the hammer differ enormously in the atmosphere precisely because terminal velocity, unlike free fall, does depend on mass.
How quickly is it reached?
Approach to terminal velocity is exponential, so strictly it is never quite reached — but it gets close fast. The exact solution is v(t) = vt· tanh(g·t / vt), which this calculator uses to report the time and distance needed to hit 95% of the final speed. For a skydiver that is roughly the first ten seconds and a few hundred metres, which is why the rest of a jump happens at a steady speed.
Where the numbers get soft
Drag coefficients are empirical, and they shift with speed, surface roughness, and orientation. Frontal area for anything that tumbles is an average rather than a constant. Air density falls with altitude, so the same object reaches a noticeably higher terminal velocity high up than at sea level. Treat the result as a good estimate rather than a precise figure.