You (77kgs) have volunteered to jump from a bridge while attached to a bungee cord. You fall 22 m from rest, beginning at time t = 0.0 s, before the cord begins to stretch. The cord stretches 14 m before you begin to move back upward. As you bounce up and down, it occurs to you that you are on an ideal bungee cord, and you will continue to oscillate until you or someone else figures out how to rescue you. What is the spring constant of the cord? I know I figured this out. TE = PE = 1/2 kdY^2 = 1/2 mV^2 = mgS = PE = TE; we solve for k = 2mgS/dY^2 = 2*77*9.8*22/14^2 = 169.4 N/M What is the time required for you to first reach (momentarily) zero velocity? A) 1.06 s B) 1.27 s C) 1.35 s D) 1.42 s E) 1.51 s Heres my work for this second part: Total time T = tf + tb; tf = sqrt(2S/g) = sqrt(44/9.8) = 2.1189sec. So it takes 2.1189 sec for the jumper to reach the point where the cord even begins to stretch. So how could any of the answers be correct???