posted by LEON on .
A bungee jumper (m = 61.00 kg) tied to a 45.00 m cord, leaps off a 75.00 m tall bridge. He falls to 5.00 m above the water before the bungee cord pulls him back up. What is the magnitude of the impulse exerted on the bungee jumper as the cord stretches?
Can anyone show me how to do this step by step?
First, you need to find the velocity when the cord begins to stretch, i.e after the first 45m of free fall
Vf^2 = Vi^2 + 2ad
Vf^2 = (0) +(2)(-9.81)(-45)
Vf = 29.7m/s
Using this velocity you can find the momentum, and since Impulse in just the chanage in momentum, it will be the momentum at this point.
I = mv
I = (61kg)(29.7m/s)
I = 1818 kg m/s