Calculating Impulse on Bungee Jumper from 78m Fall

[Punchlinegirl]

A bungee jumper (m = 75.00 kg) tied to a 48.00 m cord, leaps off a 78.00 m tall bridge. He falls to 8.00 m above the water before the bungee cord pulls him back up. What size impulse is exerted on the bungee jumper while the cord stretches?
I found the velocity by
v_final^2= V_inital + 2ad
V^2= 2(9.8)(30)
V=24.2 m/s
Impulse= change in momentum.
I think the initial momentum is 0 since he isn't moving.
Impulse= m*v
= 1818.7 kg*m/s

This wasn't right.. can someone help me out?
Thanks

 

[Diane_]

You solved for the final speed by treating it as a free fall problem. If you think about it, that's not going to be exactly true. As the bungee cord extends, it will exert a force pulling him to a stop. Try it using energy conservation - that won't depend on the acceleration remaining constant.

 

[Punchlinegirl]

Ok I tried using
mgh= (1/2)mv^2 + mgh
(75)(9.8)(78)= (1/2)(75)v^2 + (75)(9.8)(8)
Solving for v gave me 37.0 m/s..
Am I doing this right?

 

[Diane_]

Much better.

Do you know where to go from there?

 

[Punchlinegirl]

I know that impulse is the change in momentum
So
p_final-p_inital= mv_final-mv_initial.
I think the initial velocity is 0 since he's at the top of the bridge. So it would just be (75)(37.0)= 2775 kg*m/s?

 

[Leong]

the phrase 'while the cord strecthes', does that mean that u is the velocity of the man when the cord started to strectch, when he is 48 m from the bridge, and v refers to his velocity when the cord strecthed the maximum when he was 8 m above the water, v is zero at this moment since he would stop for a while before being pulled back by the cord.

 

[Punchlinegirl]

So should I use conservation of energy to find the velocity when he was at the 8 m above the water,
mgh=(1/2)mv^2+ mgh
(75)(9.8)(78)=(1/2)(75)v^2 + (75)(9.8)(30)?
I'm a little confused by the whole bungee cord stretching thing.

 

[Diane_]

According to the problem, his velocity 8 meters above the water should be 0. The bungee cord will have stopped him at that point. Think of it this way: at the top, when he first jumps off, he will have 0 momentum. Gravity will be doing its anthropomorphized best to give him some as he falls. If the cord were not there, he would have, say, x momentum at that point 8 meters above the river.

Because the cord is there, though, his actual momentum at that point will be 0. Momentum is conserved - it has to have gone somewhere. What's happened is that the cord has exerted a force as he was falling, transferring the momentum elsewhere. That force times the time it acted would give you the impulse. Conceptually, at least, that seems to me the obvious way to approach the problem. The trouble is that you know nothing about the force exerted by the bungee cord except that it's not constant, and you know nothing about the time involved. You'll have to find another approach. What's the only other approach you know for finding the impulse?

It's not unreasonable for you to be confused by the stretching of the cord. It can be analyzed, but I don't think you have the mathematics to do it right now. Follow the approach you originally outlined.

 

[Punchlinegirl]

Impulse= change in momentum= m* change in velocity
Impulse= (75)(0-37)
=-2775 kg*m/s?

 

[Leong]

Ok I tried using
mgh= (1/2)mv^2 + mgh
(75)(9.8)(78)= (1/2)(75)v^2 + (75)(9.8)(8)
Solving for v gave me 37.0 m/s..
Am I doing this right?

Since the u which i will assume refer to his velocity when the cord started to strecth; then
mgh= (1/2)mu^2 + mgh
(75)(9.8)(78)= (1/2)(75)v^2 + (75)(9.8)(78-45)
u=29.7 m/s
Impulse = 75*(0-29.7)=2227 kgm/s