# Physics Problem(Bungy Jump)

1. May 12, 2012

### Tollehej

1. The problem statement, all variables and given/known data
Matt(44kg) was going to do a bungy jump(spring constant=105N/m) that is 30m long when it's unstreched.The rope is attached to a platform 65m above the ground.What is the maximum acceleration closes to the ground.
Sorry for my bad english, i was trying to translate my homework from swedish to english.

2. Relevant equations
Don't know

3. The attempt at a solution
I'm using the conservation of engery rule thing.
Ep+Ek+Es=E2p+E2k+E2s
There is no kinetical energy or spring energy in the beginning, and no potential or kinetical energy in the end => Ep=E2s => mgh=1/2*k*x^2
and we don't know the total lenght => mg(h+x)=1/2*k*x^2
and then i solve x and i get that x= about 20m
so the total lenght when the string is attached is 20+30=50m
And then i can solve the force of the spring, F=k*x F=105N/m*50m F=5250N
And then i use ƩF=m*a => G-F=m*a => (G-F)/m=a
=> (44kg*9,81m/s^2-5250N)/44kg=-109,508 m/s^2
And this can't be true!
Any ideas?

2. May 12, 2012

### Tollehej

Is this in the right section? I'm new here.

3. May 12, 2012

### PhanthomJay

During the first 30 m, the cord is slack and has no potential energy at the beginning or end.. But the jumper is in free fall, so the jumper has Kinetic Energy at 30m. Use this 30 m drop as the starting point for your energy equation

4. May 13, 2012

### Tollehej

Sorry but I don't understand why it would be so.. But later when i was trying to solve it i got an answer that is 37 m/s^2, when the person accelerates upward from the ground. I can send my equations later, if u can take a look at them?

5. May 13, 2012

### PhanthomJay

Your answer looks about right, i thought you had made an incorrect assumption when your answer was so far off, but you initial equation looks OK if h =30, and then you made a typo using x =50 when you should have used x = 20. Looks like you've corrected that.

Note that the person does not accelerate upward from the ground. The jumper never reaches it, he comes to a halt at 50 m, or at about 15 m above ground, where his upward acceleration is highest.

6. May 13, 2012

### Tollehej

Yupp, the problem was when i was going to solve the Force of the spring F=k*x
and then i took the total lenght of the spring when it was streched but i was suppose to only the lenght that was streched which is kinda obvious! But thank you for your help!

7. May 15, 2012

### stef6987

use the 30m drop as your starting point, easiest option.

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