Question

You jump out of an airplane (with a parachute). Compute your terminal velocity (before you hit the ground). Is this dominated by the linear or quadratic air resistance term?

Answer 1

let the drag be given by
D = kv + cv^2
then

from force balance

mg - D = mdv/dt
mg - (kv + cv^2) = mdv/dt
dt/m = dv/(mg - (kv + cv^2))
integrating from vi to vf
ti = 0 to t = tf

t= -mln(|2cv - sqrt(4cgm + k^2) + k|/|2cv + sqrt(4cgm + k^2) + k|)/(sqrt(4cgm + k^2))
2cv - sqrt(4cgm + k^2) + k = (2cv + sqrt(4cgm) + k^2) + k)exp[-sqrt(4cgm + k^2)*t/m]
v = [sqrt(4cgm + k^2) - k + (sqrt(4cgm + k^2) + k)exp[-sqrt(4cgm + k^2)*t/m]]/(2c - 2c*exp[-sqrt(4cgm + k^2)*t/m])
we can see that k is always in square form and c is not
hence the quadratic term contributes more in the drag, as v^2 is bound to have more drag then v at high c