English: Graph of the velocity versus time of a skydiver reaching terminal velocity.

The time evolution is given by ${\displaystyle v={\sqrt {\frac {2mg}{\rho AC_{d}}}}\tanh \left(t{\sqrt {\frac {g\rho AC_{d}}{2m}}}\right).}$

With ${\displaystyle m=80~\mathrm {kg} }$, ${\displaystyle g=9.8~\mathrm {m\cdot s} ^{-2}}$, ${\displaystyle \rho =1.2~\mathrm {kg\cdot m} ^{-3}}$, ${\displaystyle A=1~\mathrm {m} ^{2}}$, ${\displaystyle C_{D}=0.47.}$