Terminal Velocity

Also known as falling speed limit

v_t = \sqrt{\frac{2 m g}{\rho A C_d}}

Fluid

Worked example: 80 kg skydiver, 0.7 m², Cd 1.0 → 42.78 m/s — press Try an example to run it live, then adjust anything.

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Constant used — built into this formula, no need to enter

g_{n} = 9.80665\ \text{m/s}^{2}

Standard acceleration of gravity · exact

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Terminal Velocity explained

A falling body speeds up until drag, which grows as v², matches its weight; after that the net force is zero and the speed locks in at $v_t = \sqrt{2mg / (\rho A C_d)}$ . For a belly-to-earth skydiver — roughly 80 kg, 0.7 m² of frontal area, $C_d$ ≈ 1.0, air at 1.225 kg/m³ — that works out to about 43 m/s, near the familiar 190 km/h. Pull into a head-down dive and A collapses, pushing terminal velocity past 90 m/s; deploy a parachute and A jumps by two orders of magnitude, dropping it to a survivable 5 m/s.

Density matters as much as shape, which is why Felix Baumgartner exceeded the speed of sound in 2012 at 39 km altitude: with ρ perhaps 1% of sea-level air, $v_t$ rises roughly tenfold. The formula also explains why small animals survive falls that kill large ones — mass grows with the cube of size while area grows with the square, so a mouse's terminal velocity is a fraction of a horse's. Note that $v_t$ is an asymptote, not a speed reached at a definite moment; a skydiver is within a few percent of it after about 12 seconds.

Terminal Velocity formula

v_t = \sqrt{\frac{2 m g}{\rho A C_d}}

Where

$v_t$ = Terminal velocity (m/s)
$m$ = Mass (kg)
$\rho$ = Fluid density (kg/m³)
$A$ = Frontal area (m²)
$C_d$ = Drag coefficient

Missing one of these? Work it out first, then come back

Terminal velocity — Linear Momentum (p = mv), Power from Force and Velocity (P = Fv)
Mass — Newton's Second Law, Kinetic Energy
Fluid density — Hydrostatic Pressure (P = ρgh), Drag Force (F = ½CdρAv²)
Frontal area — Drag Force (F = ½CdρAv²), Pressure (P = F/A)
Drag coefficient — Drag Force (F = ½CdρAv²), Mole Ratio from a Balanced Equation