Given: If a body of a mass m falling from rest under the action of gravity encounters an air resistance proportional to the square of velocity, then the body's velocity t seconds in the fall satisfies
v' = g - [(kv^2)/m] where k>0
and is a constant that depends on the body's aerodynamic properties and the density of the air (assume fall is too short to be affected by changes in the density).
For a 110 lb skydiver (mg=110) and with time in seconds and distance in ft., a typical value of k is .005. What is the diver's terminal velocity?
terminal velocity = v = √((mg)/k)
The Attempt at a Solution
This is a problem from the autonomous differential equation section. The first two parts of the equation were to sketch phase lines and a typical velocity curve, which was completed. This part, the third part of the question, wants the diver's terminal velocity.
For the earlier sections, I found out that v=√((mg)/k) which is the equilibrium pt./terminal velocity. I expected the answer to be √(110/.005) but the back of the textbook said that it was √(160/.005). How did 110 become 160 for the mg section? Does it have something to do with a conversion that I needed to do and missed or did it involve solving with v''?
Thanks in advance for any help.