Terminal Velocity

Equivalent forms

mg = \tfrac{1}{2}\rho C_d A v_t^2

v_t = \sqrt{\dfrac{2mg}{\rho C_d A}}

Two everyday forces — weight and air drag — meet at one speed, and the whole fall organizes itself around it.

Symbol	Name	SI unit	Dimension	Range
$v_t$	Terminal velocityoutput Steady falling speed where drag balances weight	m/s	LT^-1	0 – 120
$m$	Mass Mass of the falling body	kg	M	0.01 – 100
$C_d$	Drag coefficient Dimensionless shape-dependent drag factor	dimensionless	1	0.1 – 1.5
$A$	Frontal area Cross-sectional area facing the flow	m^2	L^2	0.05 – 2
$ρ$	Air density Density of the surrounding fluid	kg/m^3	ML^-3	0.5 – 1.3

Unit systems

Symbol	SI	CGS	Imperial
$v_t$	m/s	cm/s	ft/s
$m$	kg	g	lb
$A$	m^2	cm^2	ft^2
$ρ$	kg/m^3	g/cm^3	lb/ft^3

Where it holds

High-Reynolds-number quadratic-drag regime (cm-scale and larger bodies in air/water). For tiny particles or very viscous fluids, linear Stokes drag applies instead and

v_t \propto m

.

Dimensional analysis

[LT^{-1}]

\surd ([M][LT^{-2}] / ([ML^{-3}][L^{2}])) = \sqrt{[L^{2}T^{-2}]} = [LT^{-1}]

Both sides are a speed, m/s.

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Discovery

Galileo Galilei / later quantified by quadratic-drag theory · 1638

Galileo argued in Two New Sciences that a falling body in a resisting medium approaches a steady speed rather than accelerating without limit. The quantitative form follows from setting the quadratic drag force ½ρC_dAv² equal to the weight mg.

Try this

Step out of a plane and you don't keep speeding up forever — air pushes back until you fall at a dead-steady speed.

Drag grows with v² until it exactly cancels gravity. Adjust mass, drag coefficient, and frontal area to see the skydiver settle onto v_terminal.

Research status: stable

Real-world applications

Parachute and ejection-seat design
Raindrop and hailstone size-vs-speed relationships
Sediment settling rates in geology and water treatment
Sky-diving and BASE-jumping safety margins

Common misconceptions

Heavy objects always fall faster — only true with drag; in vacuum all masses fall identically
Terminal velocity is a property of the object alone — it also depends on air density, hence altitude
You feel a jerk 'up' at terminal velocity — acceleration smoothly goes to zero; you simply stop speeding up

Experimental verification

Skydivers measure

\sim 55\,\mathrm{m/s}

belly-to-earth

and \sim 5\,\mathrm{m/s}

under canopy, matching the formula; Felix Baumgartner's 2012 jump

hit \sim 377\,\mathrm{m/s}

in the thin upper atmosphere where

\rho is

tiny; falling-ball viscometers use the analogous Stokes result.

Derivation

Falling with quadratic drag:

m dv/dt = mg - \tfrac{1}{2}\rho C_dAv^{2}

.

At terminal velocity

dv/dt = 0

, so

mg = \tfrac{1}{2}\rho C_dAv_t^{2}

.

Solving for v_t gives

v_t = \sqrt{2mg / \rho C_dA}

.

The full solution

v(t) = v_t \tanh (gt/v_t)

shows the speed rising and saturating at v_t.

Limiting cases

A \to

large (open parachute)⟶ v_t smallBig area means drag matches weight at a gentle

\sim 5\,\mathrm{m/s}

— a survivable landing.

\rho \to 0

(vacuum)⟶

v_t \to \infty

No air, no drag: the body accelerates forever at g, never reaching terminal speed.

m \to

large, A fixed⟶ v_t grows

as \sqrt{m}

Heavier objects of the same shape fall faster — a steel ball beats a beach ball.

What if…

What if you dive head-down instead of belly-flat?

A and Cd drop, so v_t climbs $to \sim 90\,\mathrm{m/s}$ — how skydivers control fall rate.

What if you fell on the Moon?

No atmosphere, $so \rho = 0$ and there is no terminal velocity — you accelerate the whole way down.

What if the object is a tiny mist droplet?

Drag becomes linear (Stokes), and $v_t \propto m$ instead $of \sqrt{m}$ — droplets nearly hang in the air.

1

Belly-to-earth skydiver

Given ·

m:: 75
C d:: 1
A:: 0.7
ρ:: 1.225

Find · v_t

Steps

Weight $mg = 735\,\mathrm{N}$
Denominator $\rho C_dA = 0.8575$
$v_t = \sqrt{1470/0.8575} = \sqrt{1714} \approx 41.4\,\mathrm{m/s} (\sim 149\,\mathrm{km/h})$

Result ·

v_t = \surd (2\times 75\times 9.8 / (1.225\times 1\times 0.7)) \approx 41.4\,\mathrm{m/s}

2

Open parachute

Given ·

m:: 75
C d:: 1.4
A:: 25
ρ:: 1.225

Find · v_t

Steps

Same weight 735 N
$\rho C_dA = 42.9$
$v_t = \sqrt{1470/42.9} \approx 5.9\,\mathrm{m/s}$ — a safe landing speed

Result ·

v_t = \surd (1470 / (1.225\times 1.4\times 25)) \approx 5.9\,\mathrm{m/s}

Free Fall Velocity→Kinetic Friction Force→Newton's Second Law→Newton's Law of Universal Gravitation→