A sky diver weighing 180 lb falls vertically downward from an altitude of 5000ft and opens the parachute after 10 s of free fall. assume that the force of air resistance is 0.75(v) when the parachute is closed and 12(v) when the parachute is open, where the velocity v is measured in ft/s. a) find speed of the sky diver when the parachute opens b) find the distance fallen before the parachute opens c)What is the limiting velocity V_L after the parachute opens? d) Determine how long the sky diver is in the air after the parachute opens I have tried to set it up like - (dv/dt)= (.75(v))/180-a and seperating and solving but answer doesnt make sense and V=32(t)-(.75v/180)t that didnt make sense either I know I am probably setting this up wrong, I was just trying to do F=ma for the set up? please point me in the right direction. Thanks!