# Terminal Velocity Derivation

Terminal velocity is defined as the highest velocity attained by an object that is falling through a fluid. It is observed when the sum of drag force and buoyancy is equal to the downward gravity force that is acting on the object. The acceleration of the object is zero as the net force acting on the object is zero.

In fluid mechanics, for an object to attain its terminal velocity should have a constant speed against the force exerted by the fluid through which it is moving. Terminal velocity is applicable for skydiving.

The mathematical representation of terminal velocity is:

$v_{t}=\sqrt{\frac{2mg}{\rho AC_{d}}}$

Where,

vt: terminal velocity

m: mass of the falling object

g: acceleration due to gravity

Cd: drag coefficient

𝜌: density of the fluid through which the object is falling

A: area projected by the object

## Terminal velocity derivation

Deriving terminal velocity using mathematical terms according to the drag equation as follows:

$F=bv^{2}$ (drag force)

Where,

b: constant depending on the type of drag

$\sum F=ma$ (free fall of an object)

$mg-bv^{2}=ma$ (assuming that the free fall is happening in positive direction)

$mg-bv^{2}=m\frac{dv}{dt}$

$\frac{1}{m}dt=\frac{dv}{mg-bv^{2}}$ (differential form of the equations)

$\int \frac{1}{m}dt=\int \frac{dv}{mg-bv^{2}}$ (integrating the equations)

$\int \frac{dv}{mg-bv^{2}}=\frac{1}{b}\int \frac{dv}{\alpha ^{2}-v^{2}}$

Where,

$\alpha =\sqrt{\frac{mg}{b}}$

$dv=\alpha sech^{2}(\Theta )d\Theta$ (after substituting for $v=\alpha tanh(\Theta )$)

$v^{2}=\alpha ^{2}tanh^{2}(\Theta )$

After integration

$\frac{1}{b}\int \frac{\alpha sech^{2}(\Theta )d\Theta }{\alpha ^{2}-\alpha ^{2}tanh^{2}(\Theta )}$

$\frac{1}{b}\int \frac{\alpha sech^{2}(\Theta )d\Theta }{\alpha ^{2}(1-tan^{2}\Theta )}$

$\frac{1}{b}\int \frac{\alpha sech^{2}(\Theta )d\Theta }{\alpha ^{2}sech^{2}(\Theta )}=\frac{1}{\alpha b}\int d\Theta =\frac{1}{\alpha b}arctanh(\frac{v}{\alpha })+C$ (using the identity $1-tanh^{2}(\Theta )=sech^{2}(\Theta )$)

$\frac{1}{m}t=\frac{1}{\alpha b}arctanh(\frac{v}{\alpha })+C$ (from original equation)

$v(t)=\alpha tanh(\frac{\alpha b}{m}t+arctanh(\frac{v_{0}}{\alpha }))$

By substituting for $\alpha =\sqrt{\frac{mg}{b}}$

$v(t)=\alpha tanh(t\sqrt{\frac{bg}{m}}+arctanh(\frac{v_{0}}{\alpha }))$

$v(t)=\sqrt{\frac{mg}{b}}tanh(t\sqrt{\frac{bg}{m}}+arctanh(v_{0}\sqrt{\frac{b}{mg}}))$

After substituting for vt

$\lim_{t\rightarrow \infty }v(t)=\lim_{t\rightarrow \infty }(\sqrt{\frac{mg}{b}}tanh(t\sqrt{\frac{bg}{m}}+arctanh(v_{0}\sqrt{\frac{b}{mg}})))=\sqrt{\frac{mg}{b}}$

$∴ v_{t}=\sqrt{\frac{2mg}{\rho AC_{d}}}$

Therefore, above is the derivation of terminal velocity.

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