"An 80.0 kg skydiver (initially at rest) jumps out or an airplane at an altitude of 1000.0 m and opens the parachute at some altitude. Assuming that the totalretarding force on the skydiver is constant at 50.0 N with the parachute closed and the constant at 3600.0 N with the parachute open, at what altitude should the skydiver open the parachute so that he lands with a velocity of 5.00 m/s on the ground?"
Quite a mouth full.
Vf^2 = Vi^2 + 2ax
Sigma F= MA
The Attempt at a Solution
Okay im really kinda stuck on this one.
I tried figuring out the acceleration by doing:
784.8 N -3600N = 80 kg(a)
but that yields an acceleration of -35 m/s^2, And that can't be right.
I also drew a free body diagram but i can't post that up.
I know that the correct answer is 207 m because the teacher gave that to us.
Can anyone help me?