Terminal Velocity Calculator

terminal velocity = √( 2 × mass × g ÷ ( fluid density × frontal area × drag coefficient ) ) drag force = mass × g

m

Mass of the falling object in kilograms — the field also takes pounds, grams and tonnes

g

Standard gravity, 9.80665 m/s² — the value used throughout; the page is about falling near the Earth's surface, not on Mars or the Moon

ρ

Density of the fluid being fallen through, in kg/m³ — 1.225 for air at sea level and 15 °C, 1,025 for sea water. The field also takes g/cm³ and lb/ft³

A

Frontal area in square metres — the silhouette the object presents to the flow, which for a person is the whole difference between 122 mph belly-down and 211 mph head-down

Cd

Drag coefficient, dimensionless — how blunt the shape is. Roughly 1.28 for a flat plate facing the flow and 0.3 for a car; the reference table gives six shapes

Use this page whenever something falls far enough that the air matters — a skydiver, a parachute, a hailstone, a drone, a dropped tool, a ball. It answers the two questions the free-fall formula cannot: how fast the object actually gets, and whether that is fast enough to be dangerous. The free-fall formula keeps predicting more speed forever, which stops being true after a few seconds in air; this page gives the ceiling instead. Two habits make the answers more useful. First, get the frontal area right, because it is the input people guess worst and the one the speed depends on most sensitively: a person changes it by a factor of three just by changing posture, and that is a factor of 1.7 in speed. Second, treat the drag coefficient as the order-of-magnitude number it is, and reach for the reference table rather than a remembered decimal — a sphere's Cd moves from 0.47 to about 0.1 with the Reynolds number, which is exactly the effect a golf ball's dimples exploit.

1. A 200 lb skydiver, belly to earth

   1. 200 lb = 90.718 kg; 7.5 sq ft = 0.6968 m² (the area field takes square metres)
   2. Drag force at terminal velocity: 90.718 × 9.80665 = 889.64 N, which is simply the weight — that is what terminal velocity means
   3. Denominator: 1.225 × 0.6968 × 0.7 = 0.5975
   4. Numerator: 2 × 889.64 = 1,779.29
   5. Terminal velocity: √(1,779.29 ÷ 0.5975) = √2,978 = 54.6 m/s
   6. In the other units: 196.5 km/h, or 122.1 mph

   This is the number every skydiver eventually learns: about 120 mph, belly to earth, in a normal arch. Note what the third step actually says — the drag force equals the weight, because that is the definition of the terminal state. The page prints it separately precisely so you can see the balance rather than take it on trust. Also note that the mass cancels in an interesting way: a heavier skydiver falls faster, but only as the square root, so 20 percent more weight buys about 10 percent more speed.

2. The same skydiver, head-down

   1. Same mass and same drag coefficient — only the frontal area changes, from 7.5 sq ft to 2.5 sq ft
   2. 2.5 sq ft = 0.2323 m², one third of the belly-down area
   3. Denominator: 1.225 × 0.2323 × 0.7 = 0.1992
   4. Terminal velocity: √(1,779.29 ÷ 0.1992) = √8,932 = 94.5 m/s
   5. In the other units: 340.3 km/h, or 211.4 mph
   6. Drag force: unchanged at 889.64 N

   Cut the area to a third and the speed goes up by √3 = 1.73, from 122 mph to 211 mph — and that is the entire reason competitive speed skydiving exists and why the same person in jeans falls slower than the one in a tight suit. The drag force is identical in both postures, because the weight has not changed; what changed is the speed at which that force is reached. For scale, 211 mph is roughly the terminal velocity a human body reaches with no parachute at all, which is why the emergency procedure in a skydiving accident is to get flat, not to reach for anything.

3. The same skydiver under a 540 sq ft canopy

   1. 540 sq ft = 50.168 m² — this is the canopy's area, not the jumper's
   2. Drag coefficient of an open canopy: 1.42, the bluntest shape in the reference table
   3. Denominator: 1.225 × 50.168 × 1.42 = 87.28
   4. Terminal velocity: √(1,779.29 ÷ 87.28) = √20.39 = 4.5 m/s
   5. In the other units: 16.3 km/h, or 10.1 mph
   6. Descent from 3,000 ft at that rate: about 3.4 minutes

   The whole page in one comparison: the same person, the same weight, a drag force that has not moved at all, and a descent speed 12 times slower. Nothing about the jumper changed; the area went up by 72 times and the coefficient by a factor of two, and the speed went as the square root of that. The 10 mph figure is the one to sanity-check against experience — a parachute descent is a step off a high wall, which is why landing technique matters at all and why a canopy that fails to open fully is survivable but a hard landing is not.

The formula assumes the drag force grows with the square of the speed, which is a good approximation at high Reynolds numbers and a poor one at low: for a very small or very slow object — a dust particle, a settling cell, a balloon — drag is closer to linear in speed and the real terminal velocity will be different from what this page says. The drag coefficient is not a constant even for a fixed shape; it depends on the Reynolds number, so the sphere's 0.47 above is the value for a smooth sphere over a particular range and drops towards 0.1 at higher speeds, which is the effect a golf ball's dimples deliberately induce. The page adds nothing about the approach to terminal velocity: it gives the ceiling, not the time to reach it, and the first few seconds of a fall are genuinely described by the free-fall formula instead. It assumes the fluid is still and uniform, so wind, thermals, a fall through cloud or a descent into denser air near the ground are all outside it — the density field is a single number, not a profile. It assumes the object does not tumble, rotate or change shape, which for a human means assuming a held posture; a body that starts to spin is a different problem. It ignores buoyancy, which matters for anything light enough to float in the fluid, and it ignores the compressibility of the air at speeds above about 100 m/s, where the numbers here begin to overstate the drag. Finally, it is a single-object calculation: a parachute with a payload, a cluster of objects on one line, or anything with a tether is a system and not a body.

What is the terminal velocity formula?

v = √(2mg ÷ ρACd). The numerator is twice the weight; the denominator is the fluid density times the frontal area times the drag coefficient, all three of which slow the fall. For an 80 kg skydiver belly to earth — 0.7 m² and Cd 0.7 in air at 1.225 kg/m³ — that works out to 51.1 m/s, which is 114 mph. The square root is why doubling the mass only gives about 1.41 times the speed.

How fast does a skydiver fall?

About 120 mph belly to earth, in a normal arch, and about 210 mph head-down with the arms and legs tucked. Both come from the same formula and the same weight: the only thing that changes between them is the frontal area, from 7.5 square feet to 2.5. Under a 540 square foot canopy the same jumper descends at about 10 mph, which is a step off a wall rather than a fall.

Does a heavier object fall faster?

It reaches a higher terminal velocity, but only as the square root of the mass, so the effect is much weaker than people expect. Going from 80 kg to 90.7 kg — 13 percent heavier — raises the belly-to-earth speed from 51.1 to 54.6 m/s, about 7 percent. In a vacuum there is no difference at all: without air, everything falls at the same rate. In air the difference is real but modest, and shape matters far more than weight.

What drag coefficient should I use?

Take it from the reference table rather than from memory, and treat it as a range rather than a decimal: 1.28 for a flat plate facing the flow, 1.42 for an open parachute canopy, 1.2 for a long cylinder across the flow, 0.7 for a person belly to earth, 0.47 for a smooth sphere and 0.3 for a car. If you are between two rows, pick the blunter one — Cd falls as a shape becomes more streamlined, and guessing too high keeps the answer conservative.

How is terminal velocity different from free fall speed?

Free fall is what happens before the air matters, and it has no speed limit — the formula v = gt keeps predicting more speed forever. Terminal velocity is the ceiling that drag imposes, reached when the drag force grows to equal the object's weight. For a person that happens after about 10 to 12 seconds and roughly 1,500 ft of falling, after which the speed stops changing. Below about 5 m/s, or for objects smaller than a grain of sand, the quadratic drag law this page uses is the wrong model.

What happens if I fall into water instead of air?

The fluid density goes from 1.225 kg/m³ to about 1,025, roughly 840 times larger, and the terminal velocity falls by the square root of that — a person reaches about 1.8 m/s in sea water rather than 51 m/s. That is the honest reading of the formula and it is also why falling into water from a great height is still fatal: you reach the surface at the air terminal velocity first, and the water then has to stop you in a fraction of a second.

• Drag Forces (College Physics 2e, §5.2) — the quadratic drag law, the drag coefficient, and terminal velocity as the speed where drag equals weight — OpenStax
• Drag coefficient of common shapes — the reference values for a flat plate, a sphere, a cylinder and a car body that the table below is drawn from — Wikipedia
• NIST Guide to the SI — standard gravity as 9.80665 m/s², the density of air and water in kg/m³, and the exact factors behind the square-foot and pound conversions used above — National Institute of Standards and Technology