1. The problem statement, all variables and given/known data A mountain climber is attached to a rope. She slips, and after she has fallen 4.5 m, the rope begins to help decelerate her. If the constant deceleration is 5.2g, how many cm will the rope stretch before she stops? (For most real ropes, the deceleration depends on the stretching of the rope.) 2. Relevant equations x = x0 + v0t + 1/2 at2 v = v0 + at 2(x-x0) = (v+v0)t v^2 = v^02 + 2a(x − x0) g = 9.8 m/s^2 3. The attempt at a solution mgh1 = mgh2 mass cancels on each side (9.8m/s^2) * 5.2 = 50.96 m/s^2 (50.96 m/s^2) h = 9.8m/s^2 * 4.5 meter h = 0.865 meters of rope stretch = 86.5 cm => incorrect answer. Please help!!!