Terminal Velocity Calculator: v_t = √(2mg / ρAC_d)
Terminal Velocity Calculator
Terminal velocity is the speed at which drag has grown to equal weight, so the acceleration is zero and the fall stops getting faster. Enter mass, frontal area, drag coefficient and air density, and the page gives the speed, how long it takes to get to within 5% of it, and how far you fall doing so — with the whole speed–time curve drawn to your own numbers.
Terminal speed, the approach to it, and the drag force on the way
An 85 kg skydiver falls belly to earth, presenting a frontal area of 0.70 m² with a drag coefficient of 1.0, through air of density 1.21 kg/m³. Take g = 9.81 m/s². Find the terminal velocity, and how long it takes to get close to it.
Set drag equal to weight and solve
- C_d
- drag coefficient, a dimensionless number for the shape. It depends on the shape and on the flow, not on the size.
- ρ
- density of the fluid being pushed aside, in kg/m³ — the air, not the object.
- A
- projected frontal area in m², the area the flow sees.
- v
- speed through the fluid. Drag goes as v², which is why terminal velocity exists at all: a force that grows with the square of speed will always catch up with a constant weight.
- v_t
- terminal velocity, where the two balance. Note the square root: every quantity inside it has only half the leverage you would expect.
Worked example
An 85 kg skydiver falls belly to earth, presenting a frontal area of 0.70 m² with a drag coefficient of 1.0, through air of density 1.21 kg/m³. Take g = 9.81 m/s². Find the terminal velocity, and how long it takes to get close to it.
Write down the balance, which is the whole derivation. At terminal velocity nothing is accelerating, so drag equals weight: ½CdρAv² = mg. Everything else is algebra.
Rearrange. vt = √(2mg / ρACd). Top: 2 × 85 × 9.81 = 1667.7. Bottom: 1.21 × 0.70 × 1.0 = 0.847. Divide: 1968.9. Square root: 44.37 m/s, which is 159.7 km/h. OpenStax works the same case with g = 9.8 and quotes about 44 m/s, so the two agree.
Check the drag force. ½ × 1.0 × 1.21 × 0.70 × 44.37² = 0.4235 × 1968.9 = 833.9 N, and the weight is 85 × 9.81 = 833.9 N. They match because that is what terminal velocity means, and it is the fastest check available on this kind of question.
How long does it take? Solving the motion gives v = vt tanh(gt/vt), which approaches vt but never quite arrives. The natural time is vt/g = 44.37/9.81 = 4.52 s, and reaching 95% of terminal velocity takes 1.832 of those, so 8.29 s, by which point the diver has fallen 234 m.
Now dive head first and watch the square root work. The area drops to about 0.18 m² and the drag coefficient to 0.70, so the denominator falls by a factor of 5.56 and the speed rises by √5.56 = 2.36 — to about 105 m/s. OpenStax's own head-first example, a 75 kg diver at the same area and drag coefficient with g = 9.8, comes out at 98 m/s.
Where this goes wrong. Forgetting the square root, and using total surface area instead of frontal area, are the two reliable errors. A third is assuming a heavier object falls faster in general — it does reach a higher terminal speed, but only as √m, and in a vacuum it would fall at exactly the same rate as a light one.
Change one thing and the terminal speed moves by the square root of it
| What you change | Factor on v_t | Numerically |
|---|---|---|
| Double the mass | √2 | × 1.414 — up 41% |
| Halve the mass | 1/√2 | × 0.707 — down 29% |
| Double the frontal area | 1/√2 | × 0.707 — down 29% |
| Open a parachute, area × 25 | 1/5 | × 0.200 — down 80% |
| Double the drag coefficient | 1/√2 | × 0.707 |
| Go to air of half the density | √2 | × 1.414 — faster, in thinner air |
| Go from air to water, ρ × 826 | 1/√826 | × 0.0348 — about 29 times slower |
| Triple the mass and triple the area together | 1 | No change at all |
Drag coefficients, quoted from the two sources this page uses
| Shape | C_d | Source |
|---|---|---|
| Typical aerofoil | 0.05 | OpenStax Table 6.2 |
| Bullet | 0.295 | NASA Glenn, Shape Effects on Drag |
| Sphere | 0.45 | OpenStax Table 6.2 (NASA gives 0.07 to 0.5, flow-dependent) |
| Skydiver, feet first | 0.70 | OpenStax Table 6.2 |
| Bicycle | 0.90 | OpenStax Table 6.2 |
| Skydiver, horizontal | 1.0 | OpenStax Table 6.2 |
| Circular flat plate | 1.12 | OpenStax Table 6.2 |
| Flat plate | 1.28 | NASA Glenn, Shape Effects on Drag |
| Raindrop | not published in either source | This page uses the sphere value and says so |
Why a falling object stops accelerating, and why the square root changes everything
Nothing falls for ever at increasing speed in air, because drag grows with the square of the speed and weight does not grow at all. Release an object and at first the only appreciable force is its weight, so it accelerates at g. As it speeds up the drag force climbs, the net force shrinks, and the acceleration falls. When drag has grown to exactly match the weight the net force is zero, the acceleration is zero, and the speed stops changing. That speed is the terminal velocity, and the condition that defines it — drag equals weight — is the only thing you need to derive the formula.
Notice what terminal velocity is not. It is not a maximum speed imposed by the air; a diver who dives head first falls much faster than one spread flat, and nothing stops them. It is not reached suddenly; the approach is asymptotic, so strictly it is never reached at all. And it is not a state in which the forces have disappeared — at terminal velocity the drag force is at its largest, exactly equal to the weight. The figures on this page print both so that equality is visible.
The square root is where the physics becomes surprising. Every quantity sits under it, so each one has only half the leverage you would guess. Double the mass and the terminal speed rises by 41%, not 100%. Double the frontal area and it falls by 29%, not 50%. Open a parachute that multiplies the area by 25 and the speed falls by a factor of 5 — which is the whole design problem of parachutes: area is expensive and the return on it is a square root. A canopy that cut a 50 m/s fall to 5 m/s would need a hundred times the diver’s frontal area.
The m/A grouping explains one of the best-known facts in biology. Terminal velocity depends on mass divided by frontal area, and when an animal is scaled up, the mass goes as the cube of its length while the frontal area goes only as the square. So m/A grows with length, and bigger animals have higher terminal velocities. A mouse’s terminal speed is low enough that it walks away from a fall down a mine shaft; a horse’s is not. The same scaling is why dust hangs in the air and gravel does not.
Getting there takes time, and the page gives you that too. Solving the equation of motion with quadratic drag gives v = v_t tanh(gt/v_t), a curve that leaves the origin along the no-drag straight line of slope g and bends over towards the asymptote. The natural timescale is v_t/g — 4.5 s for the default skydiver — and reaching 95% of terminal velocity takes about 1.83 of those, around 8.3 seconds and 234 metres of fall. That matters for exam questions about short drops: an object falling two metres has not got anywhere near terminal velocity, and the no-drag free fall equations are then a perfectly good model.
Where this goes wrong is the area. Both sources behind this page define the drag coefficient against the projected frontal area — the silhouette the airflow sees — and using total surface area instead inflates A by a large and object-dependent factor. For a sphere the total surface area is four times the frontal area, which halves the terminal speed you calculate. The second error is using the density of the falling object instead of the density of the air; ρ in this formula is always the fluid being pushed aside.
Two honest limitations. The drag coefficient is not really a constant: it depends on the flow, which is why NASA quotes a sphere as anything from 0.07 to 0.5 depending on conditions, and a single school figure is a reasonable average rather than a measurement of your particular object. And air density falls with altitude, so a terminal velocity worked out at sea level is the slowest that object will manage — the record free-fall speeds were all set in very thin air high above the ground, where the same person falling the same way reaches several times the sea-level figure before the thickening air slows them again.
Frequently asked questions
Why does a heavier object have a higher terminal velocity if everything falls at the same rate?
Both statements are true and they are about different things. In a vacuum everything falls at the same rate, because the extra weight of a heavier object is exactly matched by its extra inertia. In air, the drag force depends on size and shape rather than on mass, so a heavier object of the same shape needs more speed before drag can match its larger weight — and so settles at a higher terminal velocity, as √m.
How long does it take to reach terminal velocity?
Strictly, for ever: the speed approaches it asymptotically and never quite arrives. In practice the useful figure is the time to reach 95%, which this page gives — about 8.3 s and 234 m of fall for the default skydiver. The natural timescale is v_t/g, and 95% takes 1.83 of those.
What area should I use?
The projected frontal area: the area of the shadow the object would cast if the airflow were a light shining straight at it. Both sources for the drag coefficients on this page define them that way, so any other choice makes the coefficient meaningless. For a sphere of radius r that is πr², not the 4πr² of its surface.
Is there a drag coefficient for a raindrop?
Not in either source this page uses. OpenStax’s table and NASA Glenn’s page both tabulate a sphere but neither gives a raindrop, so the raindrop option here is the sphere value of 0.45 and is labelled as such. A real drop flattens into a shape rather like a bun as it grows, which raises its drag coefficient above the sphere figure — by how much is something this page could not establish and therefore does not state.
Does terminal velocity depend on the height you fall from?
Not directly, but it depends on the air density, and that falls with altitude. An object starting very high up is falling through thinner air and so has a much higher terminal velocity up there; as it descends into denser air the terminal velocity drops and the object actually slows down. That is why high-altitude free-fall records reach speeds far beyond anything possible near the ground.
Why is the drag force equal to the weight at terminal velocity?
Because that is what the phrase means. Terminal velocity is the speed at which the acceleration has become zero, and by Newton’s second law zero acceleration means zero net force, so the upward drag must exactly cancel the downward weight. It gives you a free check on any answer: compute the drag at your terminal velocity and it must come out as mg.
What happens if I put water in as the fluid?
You get the terminal velocity of something sinking, and the page handles it — although for small, slow objects in a thick fluid the v² drag law stops applying and Stokes’ linear law takes over instead. Note also that this page ignores buoyancy, which is negligible in air and emphatically is not in water: the buoyancy page covers the upthrust that would have to be subtracted from the weight first.
Related calculators
References
- OpenStax, University Physics Volume 1, §6.4 “Drag Force and Terminal Speed” (CC BY 4.0). Equation 6.5 is FD = ½CρAv²; the terminal-speed result is vT = √(2mg/ρCA); Table 6.2 lists the drag coefficients this page offers.
- NASA Glenn Research Center, Beginner’s Guide to Aeronautics, “Shape Effects on Drag”. Flat plate Cd = 1.28, wedge 1.14, sphere “varies from .07 to .5”, bullet .295, typical aerofoil .045, and “the projected frontal area of each object was used as the reference area”.
- NCERT, Physics: Textbook for Class XI, Part II, Chapter 9 “Mechanical Properties of Fluids”, §9.2. Equation 9.1 defines average pressure as F/A; Equation 9.6 gives P2 − P1 = ρgh; Equation 9.7 gives P = Pa + ρgh; Equation 9.8 is the mercury barometer. Read through IIT Kanpur’s SATHEE mirror of the chapter, because ncert.nic.in refuses the file to an automated request (robots).
- NIST, CODATA Internationally Recommended 2022 Values of the Fundamental Physical Constants, ASCII table read 11 November 2026. Standard acceleration of gravity 9.806 65 m s−2, marked (exact) rather than carrying an uncertainty; Newtonian constant of gravitation 6.674 30(15)×10−11 m3 kg−1 s−2.
physics.nist.gov/cuu/Constants/Table/allascii.txt