Bungee Jumping-Tension+Air resistance with regards to max extension

[Thiras]

Homework Statement

Bungee Jumping
l=natural length of rope
x=extension of rope
y=total distance fallen
m=mass,a=acceleration,v=velocity,g=acceleration due to gravity
k=air resistance co-efficient

Given data: The rope is stretched to twice its natural length by a mass of 75kg hanging at rest from its free end

The bungee jumper jumps off a bridge, as he falls, he experiences quadratic drag (kv^2), when the rope begins to extend, he also experiences a deceleration due to the tension in the rope. Find the maximum extension of the rope (x) with regards to the mass of the jumper (m) and the natural length of the rope(l).

Previous questions were without air resistance, which i worked out fine.

Homework Equations

F=ma (a=F/m)
T=jx (Hookes Law, j not k to avoid confusion with drag)
a=(d/dx)0.5x^2
a=v(dv/dx)

The Attempt at a Solution

General thought pattern:
Find the acceleration in terms of all the variables, from the acceleration, find the velocity through whichever means possible. Equal the velocity to 0, as this is when the extension will be at a maximum. Manipulate it into the form x=something.

Therefore:
a=acceleration due to gravity-acceleration due to tension - drag
a=g-(75g/lm)x -kv^2 ((75g/lm)x was found as the deceleration due to tension in one of the previous questions)

So i now have a formula for a in terms of the extension and the velocity, the trouble comes when i attempt to substitute a for either (d/dx)0.5x^2 or v(dv/dx). As the formula contains both x and v, I am always left with trying to do either (75g/lm)x dv or kv^2 dx. The solution my teacher came up with was to treat x or v respectively as a constant(which they're not), but i don't really think that is the appropriate way to go about it.
I've also tried to use v^2 = u^2 +2as, but i then realized that only works for constant acceleration. I can calculate the velocity of the jumper when x=0, so I am fine with tension or air resistance by themselves, it is when trying to put them both together where i become stuck.
So I am really just trying to find a way to get started on this question without running into a brick wall, the only thing i can currently think of at this point is equating x to v somehow, but i have no idea if that will work or not.
Yes, this is for an assignment, I'm not asking for anyone to do it for me, I am just looking for a place to start on this little piece of hell. If more info is needed, the assignment is here: http://www.qsa.qld.edu.au/downloads/senior/snr_maths_c_***_sample_2.pdf (9th page) or just ask. Many thanks

 

[issacnewton]

link not working. please post the original question AS IT IS.

 

[Thiras]

Dynamics — Bungee jumping
New Zealand is the home of bungee jumping. One of the
major jumps is located on a bridge over the Shotover
River near Queenstown.
In this case, the bridge is 71 m above the river.
Two types of jumps are available — wet and dry. In a dry
jump, the person’s fall ends just above the water surface.
In a wet jump the person is submerged to a depth of 1 m.
Participants jump from the bridge, fastened to an elastic
rope that is adjusted to halt their descent at an
appropriate level.
The rope is specially designed and its spring constant is
known from specifications. For the purposes of the problem, we will assume that the rope is
stretched to twice its normal length by a person of mass 75 kg hanging at rest from the free
end. It is necessary to adjust the length of the rope in terms of the weight of the jumper.
1. For a person of mass m kg, calculate the depth to which a person would fall if attached
to a rope of the type described above, with length l metres. Treat the jumper as a
particle so that the height of the person can be neglected. Discuss the assumptions
made in this calculation.
2. If you were the person jumping off the 71 m attraction, find the length of rope needed
for a dry jump, where the descent is halted 1 m above the water.
3. Now find the length of rope needed for a wet jump, where the descent would end 1 m
below the surface of the water. Find the speed of entry to the water.
4. In practice, the bungee rope is attached to the ankles of the jumper. Refine the
previous model to allow for the height of the jumper and modify the earlier calculations.
Is the difference significant?
5. At present, the model does not include air resistance. Discuss the changes which
would have to be made to the model to include air resistance, which is proportional to
the velocity of the jumper. Discuss the difficulties involved with the mathematics of this
model.
6. Read the newspaper article entitled “Bungee jumping requires leap of faith” (available
from the Bungee.com website, www.bungee.com/bzapp/press/lj.html). Use
mathematics to support or refute the journalist’s comments.

It is Q5 i am talking about

 

[Thiras]

I was skeptical as to what Q5 actually involved, but the teacher said i need to get a function x in terms of l and m
Also, i know it says proportional, not quadratic drag, teacher said to change it

 

[issacnewton]

thiras, in Q 5 they are saying that air resistance is proportional to the velocity of the jumper. why are you using quadratic terms in velocity for air resistance ?

 

[Thiras]

My teacher said that she thought that quadratic was easier to work with, and that's what she used in her workings. This might be a good time to point out that as of this morning she hasnt figured out how to do it yet, which is why I'm here asking you guys

 

[issacnewton]

nice article I found on physics involved, though it neglects air resistance
http://www.real-world-physics-problems.com/physics-of-bungee-jumping.html

here is MATLAB analysis with air drag

http://inference.us/SolutionPlatform/Documents/Matlab/MATLAB%20Results%20Document%20%20(Bungee%20Jumping).pdf