## 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 {jatex options:inline}v{/jatex} 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 {jatex options:inline}100v \times.129 =129v{/jatex} kg of air acquires a speed {jatex options:inline}v{/jatex} and momentum {jatex options:inline}129v \times v =129v^2 {/jatex} kg m/s.

According to Newton's Second Law, Force = Rate of Change of Momentum.

The parachutist experiences an upwards force {jatex options:inline}129v^2{/jatex} N and downwards force due due his weigh {jatex options:inline}mg{/jatex}.

Hence {jatex options:inline}129v^2 =mg \rightarrow v = \sqrt{\frac{mg}{129}}{/jatex}

For a parachutist of mass 100 kg {jatex options:inline} v = \sqrt{\frac{70 \times 9.81}{129}}=2.31 {/jatex} m/s.