Bungee jump?

Question

A bungee jumper (m = 80.00 kg) tied to a 36.00 m cord, leaps off a 66.00 m tall bridge. She falls to 5.00 m above the water before the bungee cord pulls her back up. What size impulse is exerted on the bungee jumper while the cord stretches?

Answer 1

The jumper travelled 66-36 m before the cord started stretching so she was basically 30 m above the ground. that means the cord stretched 25 m. her initial velocity after falling 36 m was
vf^2 - vi^2 = 2*a*x
vi=0, a=-9.8, x = -36
vf = 26.56 m/s
then she stops at 5 m which means she stretched 25 m as said before. therefore, impulse is change of momentum over time
delta x = 0.5 (vi+vf)*t
25=0.5(26.56+0)*t
t=1.88 s
impulse = m(vf-vi)/t = 80(0-26.56)/1.88 = -1130.35 N.s

Answer 2

The easiest way to do this is to figure out the jumper's gravitational potential energy at the start. She falls a total of 66-5= 61 meters, so use 61m to determine the gravitational potential energy. Then figure out how fast she would be moving at the bottom if she didn't have the bungee cord. All her potential energy will be converted to kinetic energy. Then figure out her momentum at the bottom (without a bungee). Impulse is equal to the change in momentum when using the bungee. Since her speed is zero at the bottom, the impulse will equal the momentum she had when reaching the same height without the bungee.

ps. ANarchy is incorrect about impulse being change in momentum over time. It is only the change in momentum.