Falling Body with Air Resist

F_total = F_grav - F_air

ma = mg - kv where k units are kg/sec

mdv/dt = mg - kv

put into standard form:

mdv/dt + kv = mg

dv/dt + vk/m = g <----Standard form

find u = e^∫k/mdt = e^kt/m

multiply by u:

e^kt/mdv/dt + e^kt/mvk/m = ge^kt/m

use product rule:

d/dt(ve^kt/m) = ge^kt/m

integrate both sides:

ve^kt/m + C₁ = mg/ke^kt/m + C₂

solve for v:

v = mg/k - (C₁ - C₂)/e^kt/m

v = mg/k - Ce^-kt/m

If at time 0, the body is at rest then v(0) = 0 = mg/k - C so C = mg/k

therefore v = mg/k - mg/ke^-kt/m

or

v(t) = (mg/k)(1-e^-kt/m)

when t goes to infinity, e^-kt/m goes to 0 so the terminal velocity = mg/k

The value of k depends on mass and shape, so it can be determined by measuring terminal velocity:

average terminal velocity of an 80kg person = 148mph = 66m/s so k = 80kg(9.8m/s²)/66 m/s = 11.8 kg/sec