How this instrument works
Terminal velocity is the speed at which a falling body stops accelerating because drag has grown large enough to cancel gravity exactly. Drag force from air resistance rises with speed squared, ½ρCdAv², while weight, mg, stays fixed. Setting these equal and solving for v introduces a square root in v_t = √(2mg ⁄ (ρCdA)): heavier objects and thinner air push terminal speed up, while more drag coefficient, more frontal area, or denser air pull it down.
This second formula, v(t) = v_t·tanh(gt ⁄ v_t), describes the approach toward that ceiling, not the ceiling itself. Early on, when gt is small relative to v_t, tanh of a small number is nearly that number, so speed still tracks ordinary vacuum free fall, v = gt. As gt grows, tanh saturates toward 1 and speed levels off asymptotically, never technically reaching v_t but getting close enough within a few seconds that the difference stops being measurable with a stopwatch.
This quadratic drag model behind both formulas holds for objects moving through air at Reynolds numbers typical of skydivers, vehicles, and sports equipment — roughly walking pace and faster. It breaks down at small scale: dust, mist, and fine raindrops fall slowly enough that viscous forces dominate over inertial ones, and their terminal speed follows Stokes' law instead, where drag scales with v rather than v². This model also assumes a fixed Cd and A, true only for a body holding one rigid shape and orientation throughout the fall.
- Enter its weight into Falling object's mass, in kilograms or pounds.
- Set Drag coefficient, Cd to match body shape — about 1.0 for a belly-down human, lower for a streamlined dive.
- Enter Cross-sectional area — the surface facing oncoming air — in square metres or square feet.
- Set Air density to match altitude; 1.225 kg/m³ is sea-level standard.
- Enter Fall time in seconds, then read Terminal velocity and Velocity at the given fall time below.
Worked example — an 80 kg skydiver in belly-to-earth fall
Take an 80 kg skydiver falling belly-to-earth, drag coefficient Cd = 1.0, cross-section A = 0.7 m², and sea-level air density ρ = 1.225 kg/m³. That terminal-velocity formula gives v_t = √(2 × 80 × 9.80665 ⁄ (1.225 × 1.0 × 0.7)) = 42.7763 m/s, about 154.0 km/h — a well-known figure skydivers quote for a standard belly-down position.
After a fall time of 10 s, this second formula gives v(t) = 42.7763 × tanh(9.80665 × 10 ⁄ 42.7763) = 41.9122 m/s, about 98.0 percent of terminal velocity already. Vacuum free fall over those same 10 seconds would reach v = gt = 98.07 m/s, more than double; squared-speed growth of drag catches up with gravity's constant pull far faster than intuition built on a vacuum would suggest.
Questions
Why does the terminal velocity formula have a square root?
Because drag force grows with the square of speed, ½ρCdAv², while weight mg stays constant. Setting drag equal to weight and solving for v isolates v² first, so recovering v itself takes a square root — which is why doubling the mass only raises terminal velocity by about 41 percent, not 100 percent.
How is this different from free fall in a vacuum?
In a vacuum, speed grows without limit as v = gt — 10 seconds of fall gives 98.07 m/s no matter what's falling. With air resistance, drag climbs with v² until it cancels gravity, so speed levels off at a fixed terminal velocity instead of climbing forever; the tanh term in v(t) is what produces that flattening.
What does the drag coefficient Cd actually represent?
It is a dimensionless number describing how much of the airflow around a shape turns into drag-producing turbulence rather than sliding past smoothly. A belly-down skydiver runs close to Cd = 1.0; tucking into a head-down dive can drop it toward 0.7, which is why orientation alone changes fall speed substantially.
Does air density really change enough to matter?
Yes. Air density falls with altitude and temperature — roughly 1.225 kg/m³ at sea level but closer to 0.82 kg/m³ near 4,000 metres. Since terminal velocity scales as the inverse square root of density, thinner high-altitude air noticeably raises the speed reached before drag catches up with gravity.
Why does the fall speed curve flatten instead of keep climbing?
Because drag force keeps growing with v² as speed increases, until it exactly equals weight and net force — hence acceleration — hits zero. The tanh function in v(t) = v_t·tanh(gt ⁄ v_t) is the mathematical shape of that approach: fast rise early on, then a smooth flattening as the object nears v_t.
Can this formula be used for raindrops or falling leaves?
Only roughly. The quadratic-drag equation used here fits objects moving through air fast enough for inertial forces to dominate — skydivers, vehicles, dropped tools. Fine mist, dust, and small raindrops fall slowly enough that viscous forces dominate instead, following the linear Stokes' law regime, where this formula overstates their terminal speed.