A 72 kg bungee jumper jumps from a bridge. She is tied to a bungee cord whose unstretched length is 13 m, and falls a total of 38 m. Calculate the spring stiffness constant "k" of the bungee cord, assuming Hooke's law applies. Calculate the maximum acceleration she experiences.

Answer 1

mgh=1/2 k x^2
72(9.8)*38=1/2 k (38-13)^2
solve for k

Maximum acceleration? as she begins to fall, it is -9.8m/s^2

Now at the bottom, she is accelerating upward.

Forcebottom= -mg+k(38-13)
and Force= ma
solve for a.