Okay...I am think that I am totally lost on this problem. A helicopter holding a 70-kilogram package suspended from a rope 5.0 meters long accelerates upward at a rate of 5.2 m/s^s. When the upward velocity of the helicopter is 30 meters per second, the rope is cut and the helicopter continues to accelerate upward at 5.2 m/s^2. Determine the distance between the helicopter and the package 2.0 seconds after the rope is cut. What I did to solve it was, I solved for final velocity, using the formula Vf=Vi+a*t and I got 40.4 m/s. Then I solved for the displacement by using the formula 1/2(Vi+Vf)t and I got 70.4 m. Is any part of this right? If not, can someone point me in the right direction please!