Terminal Velocity Calculator
Calculate the terminal velocity of a falling object from its mass, drag coefficient, air density, cross-sectional area and gravity. Enter the values and get the maximum falling speed in m/s, km/h, mph and more.
Find the Maximum Falling Speed of an Object with Air Drag
Terminal Velocity Formula
Where:
- \(v_t\) — terminal velocity (m/s)
- \(m\) — mass of the object (kg)
- \(g\) — gravitational acceleration (m/s²)
- \(\rho\) — density of the fluid, usually air (kg/m³)
- \(A\) — cross-sectional (frontal) area facing the flow (m²)
- \(C_d\) — drag coefficient (dimensionless)
The formula comes from balancing two forces: the weight \(mg\) pulling the object down and the aerodynamic drag \(\tfrac{1}{2}\rho v^2 C_d A\) pushing back up. When the two are equal, the net force is zero and the object stops accelerating — that speed is the terminal velocity.
Terminal Velocity – Calculation Example
Example: an 80 kg skydiver falls belly-down through sea-level air. The frontal area is about 0.7 m², the drag coefficient for a belly-down position is roughly 1.0, and air density at sea level (15 °C) is 1.225 kg/m³.
\[ v_t = \sqrt{\frac{2 \times 80 \times 9.80665}{1.225 \times 0.7 \times 1.0}} = \sqrt{\frac{1569.06}{0.8575}} \approx 42.8 \text{ m/s} \]
Result: the terminal velocity is about 42.8 m/s, or roughly 154 km/h (95.7 mph) — right in the range expected for a belly-down skydiver. At that speed the drag force equals the weight: \(F_d = 784.5\) N, matching \(W = 80 \times 9.80665 = 784.5\) N.
How to Use the Terminal Velocity Calculator
- Enter the Mass (m) of the falling object and choose its unit (kg, g or lb).
- Enter the Drag coefficient (Cd). Use 0.47 for a sphere, about 1.0 for a belly-down skydiver, 0.7 for a head-down skydiver, or 0.04 for a streamlined body.
- Enter the Air density (ρ) in kg/m³, g/L or lb/ft³. Sea-level air at 15 °C is about 1.225 kg/m³; at 5 km altitude it drops to roughly 0.736 kg/m³.
- Enter the Cross-sectional area (A) — the frontal area facing the flow — in m², cm², ft² or in².
- Select the Gravity (g) for the location: Earth, Moon, Mars or Jupiter.
- Read the terminal velocity in m/s, km/h, mph, ft/s and knots, along with the forces and the ballistic coefficient \(m/(C_d A)\).
What Is Terminal Velocity?
When an object is dropped, gravity accelerates it downward. But air resistance (drag) grows with the square of the speed: double the speed and the drag quadruples. As the object falls faster, the drag force keeps increasing until it exactly equals the object's weight. At that moment the forces cancel, the acceleration becomes zero, and the object continues falling at a constant speed — its terminal velocity.
Terminal velocity is not a fixed number for a given object; it depends on the medium and the object's orientation. A skydiver belly-down reaches about 50 m/s (180 km/h), but by going head-down and pulling the arms in (smaller area, lower drag coefficient) the same person can exceed 90 m/s. Open a parachute and the area explodes while \(C_d\) rises, cutting the terminal speed to a survivable 5–8 m/s. A small hailstone falls at 10–30 m/s, a raindrop only about 9 m/s.
Notice what the formula tells you physically. Terminal velocity grows with the square root of mass and gravity: heavier objects or stronger gravity mean faster fall. It shrinks with the square root of density, area and drag coefficient: denser air, a bigger frontal area or a blunter shape all slow the object down. Because mass and area act in opposite directions, a dense compact object (a cannonball) falls much faster than a light spread-out one (a sheet of paper) — this is why Galileo's feather and hammer only fall together in a vacuum.
An object never exactly reaches its terminal speed; it approaches it asymptotically. In practice, a falling body gets to about 95% of \(v_t\) after roughly \(3v_t/g\) seconds, which is a few seconds for most everyday objects in air.
Assumptions, Limitations and Practical Tips
- Quadratic drag only. The formula assumes drag is proportional to \(v^2\), which holds for typical falling objects at high Reynolds numbers. Very small, slow particles (dust, mist droplets) follow Stokes (linear) drag instead and fall slower than this formula predicts.
- Constant inputs. Real air density falls with altitude — a skydiver jumping from 10 km reaches higher speeds in the thin upper air than at sea level. For high-altitude falls, run the calculation with the density at different heights and see how \(v_t\) changes.
- Rigid, fixed-orientation object. \(C_d\) and \(A\) are treated as constant. A tumbling object or a flexible body (like a person) changes both, so treat the result as a good estimate, not an exact figure.
- No buoyancy. Buoyancy is negligible for dense objects in air but matters for light objects in water; in that case subtract the buoyant force from the weight first.
- Reference values. Useful drag coefficients: sphere 0.47, cube 1.05, streamlined body 0.04, cyclist 0.9–1.1, bald eagle in a dive as low as 0.16.
If you want to analyze the fall before terminal speed is reached — distances, times and intermediate speeds — combine this result with the Kinematic Equations Solver or the Final Velocity Calculator. The energy view of the fall is covered by the Kinetic Energy and Conservation of Energy calculators, while the Friction Force Calculator handles the analogous velocity-independent resistance on surfaces.
Frequently Asked Questions
What is the terminal velocity of a human falling without a parachute?
A typical adult in a stable belly-to-earth position reaches about 50–55 m/s (180–200 km/h, roughly 120 mph). Going head-down with arms tucked can push this above 90 m/s (320 km/h). The exact value depends on body mass, clothing and position. Never rely on these figures for safety — this is physics estimation, not skydiving guidance.
How do I work out the cross-sectional area for a simple shape?
Use the frontal area facing the direction of motion. For a sphere of diameter d, it is \(A = \pi d^2/4\) — a 10 cm ball gives about 0.0079 m². For a cylinder falling flat-end first, use \(\pi r^2\); for a falling cube or box, use the area of the face pointing downward. If the object tumbles, take an average of the extreme orientations.
Does an object reach terminal velocity in a vacuum?
No. With no air there is no drag, so nothing balances gravity and the object keeps accelerating for the whole fall. This is why a feather and a hammer dropped on the Moon (shown in the Apollo 15 experiment) hit the ground at the same time. Terminal velocity only exists in a fluid such as air or water.
Can I use this calculator for objects falling through water?
Only as a rough estimate. Set the density to about 1000 kg/m³ for fresh water, but two corrections matter: buoyancy reduces the effective weight (subtract the displaced water's weight from mg), and many objects in water move at low Reynolds numbers where linear Stokes drag applies instead of quadratic drag. For a sinking ball or diver the quadratic formula is usually still a decent approximation.
How do I convert terminal velocity between m/s, km/h and mph?
Multiply m/s by 3.6 to get km/h, or by 2.237 to get mph. Going the other way, divide km/h by 3.6 or mph by 2.237. For example, 42.8 m/s is 154 km/h and about 95.7 mph. For feet per second, divide m/s by 0.3048.
Why do heavier objects fall faster in air if gravity accelerates everything equally?
Gravity does accelerate all masses equally, but air resistance does not care about mass. Doubling the mass doubles the weight while drag stays the same for the same size and shape, so a heavier object must fall faster before drag balances its weight — \(v_t\) grows with \(\sqrt{m}\). Two objects of equal mass can also fall at different speeds if one is compact (small A) and the other spread out (large A).
What drag coefficient should I use if I don't know the object's shape?
For a rough guess of an irregular blunt object, use 1.0–1.2 (similar to a cube or a person). A rounded or hemispherical shape is around 0.4–0.5, and a streamlined body like a bullet or airship hull can be as low as 0.04. Since \(v_t\) depends on the square root of \(C_d\), a twofold error in the drag coefficient only changes the result by about 40%, so a reasonable estimate is usually good enough.