## Estimating Terminal Speed of a Parachutist

Suppose a parachute has an area of 100m

^{2}. 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 m^{3}. 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.