The equation for drag force is:
$$
F_D = \frac{C_D \rho_{air} V^2 A}{2}
$$
Where \( C_D \) is the drag coefficient, \( \rho_{air} \) is the density of air ( 1.23 kg/m3 ),
\( A \) is the cross-sectional area of the object, and \( V \) is its velocity. The following system of
ordinary differential equations describes the velocity ( \( V \) ) and height ( \( Y \) ):
$$
\frac{ dY }{ dt } = - V
$$
$$
\frac{ dV }{ dt } = g - \frac{ F_{D} }{ m }
$$
$$
Y(0) = Y_{0}
$$
$$
V(0) = 0
$$
where \( m \) is the object's mass, \( g \) is gravitational acceleration (9.81 m/s2), \( t \) is
time, and \( Y_{0} \) is the initial height of the object. In this simulation, velocity and height were
calculated using Euler's Method of numerical integration, but an analytical solution exists to find velocity
\( V(t) \) and height \( Y(t) \) as functions of time:
$$
V(t) = \mathrm{ tanh } \left[ \left( \alpha g \right)^{ \frac{1}{2} } t \right] \sqrt{ \frac{ g }{ \alpha }
}
$$
$$
Y(t) = Y_0 - \frac{ \mathrm{ ln } \left[ \mathrm{ cosh } \left( g^{ \frac{1}{2} } t \right) \right] }{
\alpha }
$$
where
$$
\alpha = \frac{ A C_{D} \rho_{air} }{ 2 m }
$$
A FIFA standard soccer ball has a circumference of 0.69 ± 0.01 m and mass of 0.43 ± 0.02 kg.
The parachutist (without the parachute) was assumed to weigh 75 kg. The parachute used for reference is a
Falcon
265 parachute, which
has an area of 24.6 m
2 and mass of 5 kg.
