Suppose a parachute has an area of 100m2. As the parachutist falls, some air will be pushed down and some will be pushed aside. If we assume that air underneath the parachute, initially stationary, is pushed down with a speed equal to the speed of the parachutist, then it will gain momentum.
If the parachutist is falling with a terminal speed
\[v\]
then the volume of air displaced per second is 100v m3. The density of air decreases with increasing altitude. At sea level it is about 1.29 kg/m. If we take this as the density of air, every second \[100v \times.129 =129v\]
kg of air acquires a speed \[v\]
and momentum \[129v \times v =129v^2 \]
kg m/s.According to Newton's Second Law, Force = Rate of Change of Momentum.
The parachutist experiences an upwards force
\[129v^2\]
N and downwards force due due his weigh \[mg\]
.Hence
\[129v^2 =mg \rightarrow v = \sqrt{\frac{mg}{129}}\]
For a parachutist of mass 100 kg
\[ v = \sqrt{\frac{70 \times 9.81}{129}}=2.31 \]
m/s.