The terminal velocity of a person falling through air is about 100Km/hr. The acceleration due to gravity is 10ms-2. Use this information to find out what C is in the equation v′=C (gC-v), where g is the acceleration due to gravity and v is the velocity.

The question here is how long does it take for a falling person to reach the 90% of this terminal velocity. The computation is:

The terminal velocity vt fulfills v'=0. Therefore vt=g/c, and so c=g/vt = 10 / (100*1000/3600) = 36,000/100,000 ... / s. Incorporating the differential equation shows that the time needed to reach velocity v is

t = ln [g / (g-c*v) ] / c.

With v=.9 vt =.9 g/c,

t = ln [10] / c = 6.4 sec.