Terminal Velocity Calculator
Calculate terminal velocity from mass, drag coefficient and frontal area, with the distance and time needed to reach it.
How to use this calculator
- 1Enter the mass, then pick a shape or supply your own drag coefficient and frontal area.
- 2Adjust the fluid density if the fall is not through sea-level air.
- 3Read the terminal velocity, and the time and distance needed to approach it.
How the calculation works
v_t = √(2mg ÷ (ρ · Cd · A)) v(t) = v_t · tanh(g t ÷ v_t)- v_t
- Terminal velocity — the speed at which drag exactly balances weight, in m/s
- m
- Mass of the falling object, in kilograms
- g
- Gravitational acceleration, 9.80665 m/s² by definition at the Earth surface
- ρ
- Density of the fluid being fallen through — 1.225 kg/m³ for air at sea level
- Cd
- Drag coefficient, a dimensionless measure of how bluntly the shape meets the flow
- A
- Frontal area — the silhouette presented to the airflow, in m²
Terminal velocity comes from setting weight equal to drag: mg = ½ρv²CdA. The mass does not cancel here, unlike in free fall without air, which is exactly why a feather and a hammer fall differently in atmosphere and identically in vacuum.
Only the product Cd × A matters, not either factor alone. Halving the frontal area and halving the drag coefficient have precisely the same effect, which is why aerodynamicists quote "drag area" as a single number.
The speed grows as a hyperbolic tangent, so terminal velocity is a limit approached but never reached. 95% of it arrives quickly; the last 1% takes as long again.
Because v_t goes as the square root of mass, doubling the mass raises terminal velocity by only 41%. A heavier object of the same shape falls faster, but far less than proportionally.
Worked example
An 80 kg skydiver, belly-to-earth
- 1.Drag area: Cd 0.9 × 0.5 m² = 0.45 m².
- 2.Weight: 80 × 9.80665 = 784.5 N.
- 3.Set weight equal to drag: 784.5 = ½ × 1.225 × v² × 0.45.
- 4.Rearranged: v² = 2 × 784.5 ÷ (1.225 × 0.45) = 2,846.
- 5.v = √2,846 = 53.3 m/s, which is 192 km/h or 119 mph — the figure skydivers quote.
Result: 53.3 m/s, about 119 mph
The same jumper head-down
- 1.Head-down presents far less area: Cd 0.7 × 0.28 m² = 0.196 m².
- 2.The drag area is 2.3 times smaller than belly-to-earth.
- 3.v = √(2 × 784.5 ÷ (1.225 × 0.196)) = √6,534 = 80.8 m/s.
- 4.That is 291 km/h, or 181 mph.
- 5.Speed rises with the square root of the drag-area reduction: √2.3 = 1.52, and indeed 119 × 1.52 = 181.
Result: 80.8 m/s, about 181 mph
Why heavier things fall faster in air but not in vacuum
Galileo's result — that all objects fall at the same rate — holds only in vacuum. There, the gravitational force is proportional to mass and so is the inertia resisting it, and the mass cancels exactly.
Add air and the symmetry breaks. Drag depends on speed, shape and frontal area, but not on mass. So a heavier object of identical shape must reach a higher speed before drag grows enough to balance its greater weight. That is why the mass survives in the terminal velocity formula when it vanished from the free-fall one.
The dependence is weak, though: terminal velocity goes as the square root of mass. Double the mass and you gain only 41% more speed. Change the shape instead and the effect can be far larger — a skydiver going from belly-to-earth to head-down gains over 50% with no change in mass at all.
Drag area is one number, not two
The drag coefficient and the frontal area only ever appear multiplied together. Their product, Cd × A, is called the drag area, and it is the only aerodynamic property of the object that terminal velocity can see.
This is practically useful because Cd alone is nearly meaningless without knowing which area it was referenced to. It also explains why cyclists tuck: dropping from upright to a racing tuck cuts frontal area by roughly half, which is worth as much as halving the drag coefficient through some expensive fairing.
It is also why a parachute works. A canopy adds enormous area at a high drag coefficient, multiplying the drag area by a factor of fifty or more. Terminal velocity falls by the square root of that — around sevenfold, from roughly 120 mph to a survivable 15.
You never actually reach it
With quadratic drag the equation of motion integrates to v(t) = v_t · tanh(g t ÷ v_t), a curve that rises steeply then flattens asymptotically. Terminal velocity is a limit, not a destination.
In practice the distinction stops mattering quickly. A skydiver reaches 95% of terminal velocity in about eight seconds and some 300 metres, and the remaining 5% takes about as long again. Freefall from a typical 4,000 m exit is therefore at essentially constant speed for most of its duration, which is why the arithmetic of a jump is simpler than it looks.
One real complication this calculator does not model: air density is not constant. It falls by roughly half at 5,500 m, so a high-altitude jumper starts in thin air with a much higher terminal velocity and decelerates as they descend into thicker air. Record high-altitude jumps exceed the speed of sound for exactly this reason.
What this assumes, and where it stops
Assumptions
- Quadratic drag, which is the correct regime for objects of everyday size moving at everyday speeds through air.
- Constant fluid density over the fall, and constant gravity.
- A fixed orientation, so the drag coefficient and frontal area do not change during the fall.
- Buoyancy is ignored, which is negligible for dense objects in air but not for balloons or objects in water.
Limitations
- Air density falls with altitude — roughly halving by 5,500 m — so a high-altitude fall has a much higher terminal velocity at the start than at the end. This calculator uses one density throughout.
- Drag coefficients are approximate and vary with Reynolds number. A smooth sphere drops abruptly from about 0.47 to 0.1 at the drag crisis, which is the reason golf balls are dimpled.
- Assumes a stable orientation. Tumbling objects have a continuously varying drag area, and the effective average is difficult to predict.
- Presets are calibrated to reproduce widely quoted real-world terminal velocities rather than taken from a single published source, because published coefficient and area pairs vary substantially between textbooks.
- Ignores buoyancy, lift and any wind.
Common questions
What is the terminal velocity of a human?
About 120 mph, or 53 m/s, for a skydiver in the belly-to-earth position. Head-down it rises to roughly 180 mph because far less area meets the airflow. Under an open parachute it drops to about 15 mph, since the canopy multiplies the drag area more than fiftyfold.
Do heavier objects fall faster?
In air, yes — but much less than you would expect. Terminal velocity scales with the square root of mass, so doubling the mass raises the speed by only 41%. In a vacuum they fall identically, because the mass cancels out of the equation entirely once there is no drag to balance.
How long does it take to reach terminal velocity?
For a skydiver, about 8 seconds and 300 metres to reach 95% of it. Strictly you never reach it at all: the speed follows a hyperbolic tangent that approaches the limit asymptotically. The last few percent takes as long as the first ninety-five.
Why does the drag coefficient not matter on its own?
Because it only ever appears multiplied by the frontal area. That product, the drag area, is the single quantity terminal velocity depends on. A drag coefficient quoted without stating which area it refers to carries no information, which is why aerodynamicists prefer to quote the product.
Sources
Formula and content last reviewed on .
Results are estimates for information only, not professional advice.
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