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.