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.

answer: you will have more water.