Terminal Velocity Calculator

What terminal velocity is

Drop something and it speeds up, but only for a while. Drag grows with the square of speed, so each extra meter per second adds more resistance than the one before. At some speed the drag force equals the weight, the net force is zero, and the object stops accelerating. From then on it falls at a constant rate, which is the terminal velocity.

That is not the same as free fall in a vacuum. Without air, a feather and a hammer land together and the speed keeps climbing until they hit the ground. With air, the feather reaches its terminal speed in a fraction of a second and the hammer much later, and both stop gaining speed at a value set by their own weight and shape. NASA's beginner guide to aeronautics describes the same balance as weight equal to drag, with the object obeying Newton's first law once that happens.

The approach to the limit is gradual rather than abrupt. Speed follows a hyperbolic tangent curve, so half the terminal speed arrives quickly, 90 percent takes longer, and the last percent takes a long time. The object never fully arrives, which is why engineers talk about reaching 95 percent of terminal velocity rather than reaching it.

The formula and what each term does

Set drag equal to weight and solve for speed. The drag equation is one half times the drag coefficient times density times area times speed squared, and the weight is mass times gravity. Equating them gives the terminal velocity as the square root of 2 times mass times gravity, divided by the drag coefficient times density times area.

Because the whole right side sits under a square root, each term moves the answer by its square root rather than in proportion. Doubling the mass multiplies the speed by 1.4142. Doubling the frontal area divides it by 1.4142. A drag coefficient of 4 instead of 1 halves the speed. That square root behavior is the single most useful thing to remember about the formula.

For an 80 kg jumper with 0.7 square meters of frontal area and a drag coefficient of 1.0 in sea level air, the numerator is 2 × 80 × 9.80665 = 1569.06 and the denominator is 1.0 × 1.225 × 0.7 = 0.8575. Dividing gives 1829.8, and 1829.8 ^ 0.5 = 42.78 m/s. The weight is 784.5 N and the drag at that speed is also 784.5 N, which is the check that the answer is right.

Gravity is 9.80665 meters per second squared by definition of the standard gravity used in metrology, and it varies by about 0.5 percent between the equator and the poles. That variation is small enough to ignore for a falling object, but it is not zero, and on the Moon the same jumper would settle at 17.4 m/s because gravity is 1.62 there.

Drag coefficient and frontal area

The drag coefficient bundles everything about shape that the formula cannot see. NASA's shape effects page measured the values below in low speed wind tunnels and reports them against the projected frontal area, which is why the area you enter should be the silhouette the object presents to the airflow, not its surface area.

Area and coefficient trade against each other. A jumper who goes head down cuts the frontal area from about 0.7 square meters to roughly 0.25 and raises the coefficient to about 0.8, which together take the terminal speed from 42.78 m/s to about 80 m/s. A parachute goes the other way: 40 square meters at a coefficient of 1.5 gives 4.62 m/s, a factor of 9.26 slower than the belly to earth position.

That factor of 9.26 is worth checking against the formula: the area ratio is 40 ÷ 0.7 = 57.14 and the coefficient ratio is 1.5, so the combined ratio is 85.7, and the square root of 85.7 = 9.26. Speed falls with the square root of the drag area, not with the drag area itself.

A skydiver from 4,000 m, step by step

The reference case for this page is an 80 kg jumper in a spread-eagled position: 0.7 square meters of frontal area, drag coefficient 1.0, sea level air at 1.225 kilograms per cubic meter. The terminal speed is 42.78 m/s, which is 154.0 km/h, 95.7 mph or 140.3 ft/s.

Getting there takes time. The approach follows the hyperbolic tangent, so 95 percent of the terminal speed arrives at 1.8318 times the terminal speed divided by gravity, which is 7.99 seconds. The distance fallen by then is 1.1639 times the terminal speed squared divided by gravity, or 217 m. In the first second alone the jumper covers about 4.9 m and reaches 9.8 m/s, well short of the limit.

The 4,000 m altitude in the title matters because the air up there is thinner. Standard atmosphere density at 4,000 m is 0.8191 kilograms per cubic meter, so a jumper leaving the aircraft at that height has a local terminal speed of about 52 m/s and slows toward 42.78 m/s while descending into thicker air. The single density number in the calculator is a snapshot; a real jump moves through a range of them.

The steel ball is the case where buoyancy cannot be ignored. Its mass is 32.9 g, but the water it displaces weighs 4.2 g, so the net downward force is the weight of 28.7 g. Using the full 32.9 g would overstate the terminal speed by about 7 percent.

Air density and altitude

Density is the term that changes most between two otherwise identical falls. In the troposphere, temperature drops about 6.5 degrees Celsius per kilometer and pressure falls faster, so density falls to roughly 0.909 kilograms per cubic meter at 3,000 m and 0.5252 at 8,000 m. The NASA Earth atmosphere model is the source for the density values in the table.

The 16 percent gain from sea level to 3,000 m comes straight from the square root: the density ratio is 1.225 ÷ 0.9091 = 1.3475, and the square root of 1.3475 = 1.1607, so the speed rises by 16.1 percent to 49.65 m/s.

Water is a different order of magnitude. At 998 kilograms per cubic meter, water is about 815 times as dense as sea level air, and an object that falls at 42.78 m/s in air falls at 42.78 ÷ 28.55 = 1.50 m/s in water if the other terms stay the same. That is why a dropped phone stops moving so abruptly.

Falling in water or another fluid

The formula does not care whether the fluid is air, water or hydraulic oil. Density goes in, and the answer comes out in the same units as long as mass, area and density are consistent. A compact 5 kg object with 0.02 square meters of frontal area and a drag coefficient of 0.8 reaches 2.48 m/s in water at 20 C, which is 8.9 km/h.

Two things change in a liquid. First, buoyancy carries part of the weight, so the effective force is the object's weight minus the weight of the displaced fluid. A steel ball 2 cm across has a mass of 32.9 g but displaces 4.2 g of water, so the driving force is the weight of 28.7 g and the terminal speed comes out at 1.95 m/s rather than 2.1. Second, viscosity matters more for small particles. A 50 micrometre sand grain settles at about 2.3 mm/s, which sits inside the Stokes regime, where drag is proportional to speed rather than to speed squared and this page's formula overstates the fall rate.

The crossover is around a Reynolds number of 1. Below that, drag is linear and terminal velocity scales with the square of particle size, which is the basis of sedimentation tests and of the settling velocity used in water treatment. A 1 mm particle is already past that boundary, and above a few hundred the square law takes over and this page applies.

Where the simple model stops working

The formula treats the drag coefficient as a constant, and that is an approximation. For a sphere, NASA gives a range from 0.07 to 0.5 because the value depends on speed and surface roughness. Near the speed of sound the coefficient rises sharply, so a falling object that passes 300 m/s cannot be modeled with a single number. And a real skydiver changes shape, which changes both area and coefficient during the fall.

For the jumper, one unit of v divided by g is 42.78 ÷ 9.80665 = 4.36 seconds, and one unit of v squared over g is 186.6 m. Reading the 90 percent row gives 6.42 seconds and 155 m, and the 99 percent row gives 11.5 seconds and 365 m. The curve flattens, which is why the last few percent cost so much altitude.

Other limits are worth naming. Density changes as the object descends, so the terminal speed is a moving target. Very light objects, such as a balloon or a piece of paper, never reach the value the formula gives because buoyancy and flutter dominate. And when the fall is short, the object may never get close to terminal velocity at all: a 2 m drop reaches only about 6.3 m/s, far below the 42.78 m/s limit. The calculator reports the limit, and the time and distance rows tell you whether your fall is long enough to approach it.