What is the effective spring constant and velocity of a bungee jumper?

[toad22]

A man of mass 80,0 kg jumps out from a bridge, with a lightweight rubber rope attached to his feet. He falls 50.0 m before turning around and coming back up, which is 30,0 m further than the equilibrium point of the rubber band (without man), which we take to be like a spring when stretched (but not when squeezed; no spring force on the man until he has passed the equilibrium point.)
a) What is the effective spring constant of the rope?
b) How fast is the man going when passing the equilibrium point on his way up again?

I tried finding the spring constant like this, but I'm not sure if it's right..:
m = 80,0 kg
x = 30,0 m
F = mg
= 80,0 kg * 9,81 m/s^2
= 784 N
F = kx
k = F/x
= 784 N/30 m
= 26,1

I have no idea where to even start at problem b)..

 

[mfb]

The force at the bottom is more than 784 N - the man is slowed down and then accelerated upwards there.
What about the energy?