[SOLVED] free fall motion 1. The problem statement, all variables and given/known data A helicopter carrying Dr. Evil takes off with a constant upward acceleration of 5.0m/s^2. Secret agent Austin Powers jumps on just as the helicopter lifts off the ground. After the two men struggle for 10.0s , Powers shuts off the engine and steps out of the helicopter. Assume that the helicopter is in free fall after its engine is shut off and ignore effects of air resistance. maximum height above ground reached by the helicopter? 2. Relevant equations 1) (vy)^2 = (v0y)^2 +2ay(y-y0) 2) vy=v0y+ayt 3) y = y0 + v0yt +1/2ayt^2 3. The attempt at a solution 1) used equation 2 to derive the velocity v0y = (5m/s^2)(10s) = 50m/s plugged into equation 3 with ay = -9.8m/s^2, got: y = 0 + (50m/s)(10s) + 1/2(-9.8m/s^2)(10s)^2 = 10m wrong 2) used equation 1 with v0y = 50m/s and ay = -9.8m/s^2 and solved for y: y = (v0y)^2/(2*ay) = (50m/s)^2/(2*-9.8m/s^2) = 127.6m wrong 3) used equation 2 with ay = 5m/s^2 with v0y = 50m/s to solve for time t = v0y/g = (50m/s)/(9.8m/s^2) = 5.1s then plugged in t into equation 3 to get: y = 0 + (50m/s)(5.1s) + 1/2(-9.8m/s^2)(5.1s)^2 = 127.6m wrong again any and all help is really appreciated, I think the main problem is I have the wrong velocity but I'm not how else to find it.