1. The problem statement, all variables and given/known data A 68.0 kg bungee jumper, tied to a 31.0 m long cord, leaps off a bridge whose deck is 61.0 m above the water. He falls to 7.0 m above the water before the bungee cord pulls him back up. What is the magnitude of the impulse exerted on the bungee jumper while the cord stretches? M= 68kg Y(Length of cord) = 31m H= 61m y3(at v3=0) = 7m V3(cord fully extended)= 0 2. Relevant equations I = dP = Pfinal - Pint = m(v3-v2) y=yint+vint*t+1/2Ay*t^1/2 t=(2y)^1/2/g 3. The attempt at a solution v2 = g((2y)^1/2/g) v2 = 9.8(2(31)/9.8)^1/2=7.87m/s then i used that in conservation of energy mgh- 1/2mV^2 m's cancel out gh- 1/2V^2 9.8*61 - 1/2*7.87^2= 566.83 kg*m/s but i get Impulse is the change in momentum. You know the jumper's momentum at the end of the jump. Free fall should tell you the momentum before the bungee cord starts to pull.