Terminal Velocity Calculator
Find the fastest a falling object can go. Enter mass, cross-sectional area and shape to get terminal velocity, then see how long the object takes to reach it, how far it falls getting there, and how it compares with a skydiver, a hailstone or a dropped coin.
What a terminal velocity calculator does
Terminal velocity is the maximum constant speed an object reaches while falling when air resistance equals the force of gravity.
Calculate terminal velocity using vt = √( 2mg / (ρACd) ), where m is mass, g is gravity, ρ is air density, A is cross-sectional area, and Cd is drag coefficient.
How to use the terminal velocity calculator
- Start from a preset: Picking a familiar object fills in mass, area and shape at once, and every field stays editable afterwards.
- Enter the mass: Use whichever unit suits the object, from milligrams for a raindrop to kilograms for a skydiver. Mass has less influence than people expect, since terminal velocity depends on its square root.
- Get the area right: This is the outline the object presents to the oncoming air, and it is where most estimates go wrong. A skydiver changes it by a factor of four just by moving their arms and legs.
- Choose the shape: The drag coefficient is a single number standing in for all the messy aerodynamics. A flat plate has thirty times the drag of a streamlined teardrop of the same frontal area.
- Read the time and distance: The headline speed only matters if the fall is long enough. A dropped coin needs a few seconds and a few dozen metres; a skydiver needs about twelve seconds and a third of a kilometre.
Terminal velocity formula
A falling object is pulled down by its weight and pushed back by drag. Weight is constant, but drag grows with the square of speed, so the two must eventually match. At that moment the net force is zero, acceleration stops, and the object continues at a steady speed for the rest of the fall.
mg = ½ ρ v² A Cd
vt = √( 2mg / (ρ A Cd) )
v(t) = vt tanh( gt / vt )
d(t) = (vt² / g) ln cosh( gt / vt )
Here m is mass, g is gravitational acceleration, ρ is the density of the fluid, A is the cross-sectional area facing the flow, and Cd is the drag coefficient. The first equation is simply weight set equal to the drag equation; the second is that rearranged for speed.
The last two equations are the ones this calculator adds. They come from solving the equation of motion rather than the balance point, and they answer the question the balance point cannot: how long the object takes to get close to its terminal velocity, and how far it falls in the process. Because the hyperbolic tangent only reaches 1 at infinity, an object never truly attains terminal velocity. It reaches 95% of it after about 1.83 vt/g seconds and 99% after roughly 2.65 vt/g, which is why every honest answer here is quoted as a percentage.
Notice what is missing from the formula. Nothing about how far the object has already fallen, and nothing about how it was released. Terminal velocity is a property of the object and the fluid, not of the drop.
Derivation reference: NASA Glenn Research Center - Flight Equations with Drag, which works through the same balance of weight and drag and the resulting hyperbolic solution.
Drag coefficients by shape
The drag coefficient compresses everything complicated about a shape into one number: how the air separates behind it, how much of a wake it leaves, and how that changes with speed. These are the standard working values, and they are what the shape selector fills in for you.
| Shape | Drag coefficient | Why |
|---|---|---|
| Streamlined teardrop | 0.04 | The flow closes smoothly behind it and leaves almost no wake |
| Smooth sports ball | 0.35 | Fast enough for the boundary layer to turn turbulent and cling on |
| Hemisphere, curved side down | 0.42 | Rounded front, blunt back |
| Sphere | 0.47 | The textbook value, for moderate speeds |
| Human, head down | 0.7 | Narrow and roughly aligned with the flow |
| Long cylinder, end on | 0.82 | Sharp edges force the flow to separate early |
| Human, belly to earth | 1.0 | Broad, irregular and entirely unstreamlined |
| Cube, face on | 1.05 | Flat front with a wide separated wake |
| Flat disc or plate | 1.17 | The worst case for a solid body of its size |
| Parachute canopy | 1.4 to 1.5 | Deliberately shaped to trap and drag air along with it |
Treat these as representative rather than exact. A sphere's drag coefficient is not really a constant at all: it sits near 0.47 through a wide band of everyday speeds, then drops sharply to around 0.2 when the boundary layer becomes turbulent, which is the effect golf ball dimples exist to trigger early. A single tabulated number hides a curve.
Drag coefficient reference: NASA Glenn Research Center - Drag of a Sphere, on how the coefficient varies with Reynolds number rather than staying fixed.
Terminal velocity of common objects
All calculated with the same model, in air at sea level. The last two columns are the ones worth studying: they show how much of a fall an object needs before the headline speed means anything.
| Object | Terminal velocity | In km/h | Time to 95% | Fall needed |
|---|---|---|---|---|
| Skydiver, head down | 101 m/s | 363 km/h | 18.8 s | 1,207 m |
| Bowling ball | 76 m/s | 274 km/h | 14.2 s | 688 m |
| Skydiver, belly to earth | 43 m/s | 154 km/h | 8.0 s | 217 m |
| Golf ball | 38 m/s | 138 km/h | 7.2 s | 174 m |
| Tennis ball | 23 m/s | 83 km/h | 4.3 s | 62 m |
| Cat, legs spread | 20 m/s | 73 km/h | 3.8 s | 48 m |
| Hailstone, 2 cm | 20 m/s | 71 km/h | 3.7 s | 46 m |
| Coin | 12 m/s | 43 km/h | 2.2 s | 17 m |
| Table tennis ball | 8.3 m/s | 30 km/h | 1.5 s | 8 m |
| Raindrop, 2 mm | 6.5 m/s | 24 km/h | 1.2 s | 5 m |
| Skydiver under canopy | 6.5 m/s | 24 km/h | 1.2 s | 5 m |
Two things stand out. A raindrop is at full speed within about five metres of forming, which is why rain falls at the same rate whether the cloud is at 1 km or 4 km. A skydiver, by contrast, spends the first several seconds and a couple of hundred metres still accelerating, so the widely quoted figure of around 120 mph describes only the later part of a freefall.
The coin row is the useful one for settling arguments. A coin dropped from a tall building levels off at roughly 12 m/s, comparable to a fast throw. It stings, but the popular claim that it could be lethal does not survive the arithmetic.
Why a hammer and a feather do not land together
Galileo was right that gravity accelerates everything equally, and the Apollo 15 crew proved it on the Moon by dropping a hammer and a falcon feather side by side. On Earth they land seconds apart, and the reason is entirely contained in the terminal velocity formula.
Mass matters, but weakly
Terminal velocity rises with the square root of mass. Make an object four times heavier without changing its shape and it only falls twice as fast.
The ratio is what counts
What really decides the outcome is mass divided by area. A feather has almost no mass spread over a large outline, so drag overwhelms its weight almost immediately.
Remove the air and it vanishes
Set the density to zero and the formula divides by nothing: there is no terminal velocity in a vacuum. That is the Moon result, and it is why the Apollo demonstration worked.
The same logic explains why small animals survive falls that would kill large ones. Scale a creature down and its mass falls with the cube of its length while its area falls only with the square, so its terminal velocity drops with it.
Where you fall changes how fast you fall
Density is in the denominator of the formula, so thinner air means a higher terminal velocity for exactly the same object. The figures below are for one skydiver, belly to earth, in each environment.
| Environment | Density, kg/m³ | Gravity, m/s² | Terminal velocity |
|---|---|---|---|
| Air, sea level | 1.225 | 9.807 | 43 m/s |
| Air, 3,000 m | 0.909 | 9.797 | 50 m/s |
| Air, 10,000 m | 0.414 | 9.776 | 74 m/s |
| Air, 30,000 m | 0.018 | 9.715 | 352 m/s |
| Mars, surface | 0.020 | 3.721 | 207 m/s |
| Venus, surface | 65 | 8.87 | 5.6 m/s |
| Fresh water | 997 | 9.807 | 1.5 m/s |
The 30,000 m row is why stratospheric jumps break the sound barrier. In air that thin there is barely anything to push back, so the jumper keeps accelerating far past any speed reachable lower down, then decelerates as the atmosphere thickens beneath them. Terminal velocity is not one number for a long fall; it is a moving target the object chases all the way down.
The water row comes with a caveat. This model ignores buoyancy, which is negligible in air but dominant in a liquid: a human body is close to the density of water and will barely sink at all, while a steel ball behaves much as the formula predicts. Use liquid settings only for objects far denser than the fluid.
Atmospheric density reference: U.S. Standard Atmosphere, 1976, published jointly by NOAA, NASA and the U.S. Air Force, which is the source of the tabulated density and gravity values used here.
Where the simple formula stops working
The equation assumes one shape, one drag coefficient, one density and no buoyancy. Real falls violate all four at some point, and it is worth knowing which violation you are looking at.
| Situation | What breaks | Size of the error | What to do |
|---|---|---|---|
| Object tumbles as it falls | Area and drag coefficient both change continuously | Easily 30% either way | Use an average area, and treat the answer as a range |
| Falling through a liquid | Buoyancy is ignored, and it is large in water | Total for near-neutral objects | Only trust liquid results for objects far denser than the fluid |
| Very small or very slow objects | Drag stops scaling with the square of speed | Large for dust, mist and fog droplets | Use Stokes drag instead below about 0.1 mm |
| Speeds approaching the speed of sound | Air compresses, and the drag coefficient rises steeply | Significant above roughly 250 m/s | Treat any result near or above Mach 1 as indicative only |
| Long fall through the atmosphere | Density is not constant over the drop | Doubles between 10,000 m and sea level | Split the fall into segments and run each separately |
| Parachute opening | Area jumps by a factor of thirty in about two seconds | Speed drops roughly sevenfold | Calculate freefall and canopy phases as two separate cases |
| Sphere near the drag crisis | The coefficient itself is transitioning, not constant | Can more than halve | Check the drag coefficient at the Reynolds number you are actually at |
| Object generates lift | Not a vertical fall at all, so the balance is wrong | The model does not apply | Use a glide or projectile model instead |
Most of these push in the same direction: the real object experiences more drag than a clean textbook shape, so measured terminal velocities usually come in a little below calculated ones. The exception is the drag crisis, where a sphere can suddenly find itself with half the drag it had a moment earlier and speed up rather than settle.
Interesting Fact
In 2012 Felix Baumgartner jumped from about 39 kilometres up and reached roughly 1,358 km/h, passing through the sound barrier in freefall. He was not falling any differently from an ordinary skydiver: the air at that altitude is under a hundredth as dense as it is at sea level, so drag was almost absent and his terminal velocity was enormous. As he descended into thicker air the same formula ran in reverse and slowed him to a normal skydiving speed, without a parachute and without doing anything at all.
Frequently Asked Questions
What is terminal velocity?
It is the steady speed a falling object reaches when air resistance has grown large enough to exactly cancel its weight. The two forces sit in equilibrium, the net force is zero, so acceleration stops and the object keeps falling at a constant velocity, measured in metres per second in SI units. It is a property of the object and the fluid together, not of how high the drop started.
What is the terminal velocity of a human?
Around 53 metres per second, or about 190 km/h and 120 mph, for a skydiver lying belly to earth at typical jump altitudes. Nearer sea level, where the air density is higher, the same position gives roughly 43 m/s. Diving head down cuts the cross-sectional area to a quarter and pushes the figure past 100 m/s, and competitive speed skydivers in an extreme tucked position go well beyond 500 km/h.
How long does it take to reach terminal velocity?
Strictly it is never reached, because the approach is asymptotic. In practice, 95% of it arrives after about 1.83 times the terminal velocity divided by the gravitational acceleration. For a skydiver that is roughly eight to twelve seconds of free fall and a few hundred metres. For a raindrop it is around a second and a few metres, which is why rain arrives at the same speed regardless of cloud height.
Do heavier objects fall faster?
In a vacuum, no: gravity accelerates everything identically, so mass drops out of the equation entirely. In air, yes, but weakly. Terminal velocity grows with the square root of mass, so four times the mass gives only twice the speed, provided the shape and area stay the same. What really matters is the ratio of mass to frontal area, which is why a sheet of paper and the same sheet crumpled into a ball fall so differently.
Can a penny dropped from a skyscraper kill someone?
No. A coin is light, flat and tumbles as it falls, so the drag force overwhelms its small weight almost immediately and it levels off at somewhere around 11 to 12 m/s within the first twenty metres or so. That is comparable to a hard throw. A dense, compact projectile of the same mass would be a different matter, but a coin is the opposite of that shape. It would hurt, and nothing should be dropped from a building, but the physics does not support the popular claim.
Why does a parachute slow you down so much?
Because it attacks both terms in the denominator of the formula at once. A canopy multiplies the frontal area by roughly thirty and raises the drag coefficient from about 1.0 to about 1.5. Since terminal velocity falls with the square root of area times drag coefficient, that combination cuts the descent rate from around 50 m/s to about 6 m/s, which is a survivable landing speed. Note that it is the projected area doing the work, not the total surface area of the fabric.
Is there terminal velocity in a vacuum?
No. With no fluid there is no drag, nothing ever balances the weight, and the object accelerates for as long as it falls. Mathematically the formula divides by a fluid density of zero and returns infinity. This is why the Moon, which has no atmosphere to speak of, has no terminal velocity, and why a hammer and a feather dropped there land together.
How do I find the cross-sectional area?
Picture the shadow the object would cast with the light directly beneath it. For a sphere of diameter d it is πd²/4, or πr² if you are working from the radius, so a 2 cm hailstone gives about 3.14 cm². Do not confuse this with surface area, which for the same sphere is four times larger and plays no part in the calculation. For an irregular object, estimate the outline it presents while falling; if it tumbles, use an average of the extremes and expect the answer to carry that uncertainty with it.
Why is terminal velocity higher at altitude?
Air density sits in the denominator, so thinner air means less drag at any given speed and the object has to fall faster before the forces balance. At 10,000 m the density is about a third of its sea level value, which raises terminal velocity by roughly 70% for the same object. Above 30,000 m there is so little atmosphere left that speeds beyond Mach 1 become possible. Gravitational acceleration weakens with altitude too, but only slightly, so density is doing nearly all of the work.
How accurate is this terminal velocity calculator?
The arithmetic is exact; the inputs are not. Drag coefficients are tabulated approximations that shift with the Reynolds number, which is itself a function of speed and size, and cross-sectional area is usually an estimate. Expect to be within about 10 to 20% for a clean, non-tumbling shape, and treat comparisons between objects as far more reliable than any single absolute figure.
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Disclaimer: This terminal velocity calculator provides physics estimates for education and planning only. It applies the standard quadratic drag model, in which drag is proportional to the square of speed, with a single constant drag coefficient and a constant fluid density for the whole fall. Time and distance figures come from the analytical solution of that model.
Buoyancy is not included. This is negligible for objects falling through air but substantial in liquids, so results for water or other dense fluids are only meaningful for objects far denser than the fluid. Stokes drag governs very small or very slow objects instead, and compressibility raises the drag coefficient sharply as speeds approach Mach 1, so results near or above the speed of sound should be treated as indicative only.
Drag coefficients and densities are representative published values rather than measurements of your specific object or conditions. Tumbling, spin, lift, wind, humidity and changing air density over a long fall all shift real outcomes, sometimes substantially.
Nothing here is safety guidance. Do not use these figures to plan any activity involving falling objects, falls from height, skydiving, or dropping anything from a structure.
Last updated: August 8, 2026