The terminal velocity of an object falling towards the ground, in non-vacuum, is the speed at which the gravitational force pulling it downwards is equal and opposite to the atmospheric drag (also called air resistance) pushing it upwards. At this speed, the object ceases to accelerate downwards and falls at constant speed. An object moving downwards without power at greater than the terminal velocity (for example because it previously used power to descend, it fell from a thinner part of the atmosphere or it changed shape) will slow down until it reaches terminal velocity.

For example, the terminal velocity of a skydiver in a normal free-fall position with a closed parachute is about 195 km/h (120 Mph). It would take about 5.5 seconds to reach that speed. This speed increases to about 320 km/h (200 Mph) if the skydiver pulls in his limbs—see also freeflying. This is also the terminal velocity of the Peregrine Falcon diving down on its prey.

The reason an object reaches a terminal velocity is because the drag force resisting motion is directly proportional to the square of its speed. At low speeds the drag is much less than the gravitational force and so the object accelerates. As it speeds up the drag increases, until eventually it equals the weight. Drag also depends on the cross-sectional area. This is why things with a large surface area such as parachutes and feathers have a lower terminal velocity than small objects like bricks and cannon balls.

Mathematically, terminal velocity is described by the equation

$V\_t=\; \backslash sqrt\{\backslash frac\{2mg\}\{\backslash rho\; A\; C\_d\; \}\}$

where

$V\_t$ is the terminal velocity,

$m$ is the mass of the falling object,

$g$ is gravitational acceleration,

$C\_d$ is the drag coefficient,

$\backslash rho$ is the density of the fluid the object is falling through, and

$A$ is the object's cross-sectional area.

This equation is derived from the drag equation by setting drag equal to mg, the gravitational force on the object.

Note that the density increases with decreasing altitude, ca. 1% per 80 m (see barometric formula). Therefore, for every 160 m of falling, the "terminal" velocity decreases 1%. After reaching the local terminal velocity, while continuing the fall, speed decreases to change with the local terminal velocity.

Approximation

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Approximating terminal velocity is much more easily done than calculating the terminal velocity because of the difficulty in finding the value of $C\_d$. One simple small scale method is to hang an object out of a car window by a thin string. The terminal velocity of the object is the speed of the car when the object hangs at a 45° angle. This can be easily proven mathematically because it is when the atmospheric drag (in the horizontal direction) is equal to the force of gravity. It is when air resistance and gravity is the same.

Derivation

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A falling object experiences two forces: gravitational force, and a large-velocity drag force. The addition of these two forces results in:

$F\; =\; mg\; -\; qv^2\; \backslash ,$

where

m is mass of the object

g is the acceleration due to gravity

q is $\backslash frac\{1\}\{2\}\; \backslash rho\; C\_d\; A$ from the drag equation

which, when you plug in $F\; =\; ma$ results in a differential equation:

$m\; \backslash frac\{dv\}\{dt\}\; =\; mg\; -\; q\; v^2\; \backslash ,$

Rearange to see

$\backslash frac\{dv\}\{dt\}\; +\; \backslash frac\{q\}\{m\}\; v^2\; =\; g\; \backslash ,$

The solution to this differential equation involves a hyperbolic tangent, and is

$v(t)\; =\; \backslash sqrt\{\; \backslash frac\{gm\}\{q\}\; \}\; \backslash tanh\; \backslash left(\backslash sqrt\{\backslash frac\{gq\}\{m\}\}\; t\; \backslash right)\; \backslash ,$

Plug q back in and you get the full solution:

{|cellpadding="2" style="border:2px solid #ccccff" $v(t)\; =\; \backslash sqrt\{\; \backslash frac\{2mg\}\{\backslash rho\; A\; C\_d\}\; \}\; \backslash tanh\; \backslash left(t\; \backslash sqrt\{\backslash frac\{g\; \backslash rho\; C\_d\; A\}\{2m\}\}\; \backslash right)\; \backslash ,$

The maximum value of tanh is just 1, so as time goes on the velocity of this falling object approaches a terminal velocity:

$v\_\{t\}\; =\; \backslash sqrt\{\; \backslash frac\{2mg\}\{\backslash rho\; A\; C\_d\}\; \}\; \backslash ,$

See also

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• Acceleration
• Drag (physics)
• Newton's Laws

Gravity

Velocidad terminal