1. Aug 22, 2006

### Lethal

Hey guys,

I've tried to do this quadratic question using the formula but can't seem to get it and don't know if its right or not.

x= -b+-√b^2 – 4ac / 2a

a= 37.5
b= L
c= L^2

Need to find X the sub X in to find L in metres.

2. Aug 22, 2006

### Integral

Staff Emeritus
You will have more luck getting help if you post the entire problem. As posted we do not have enough information to solve for x, or L.

3. Aug 22, 2006

### Lethal

here is the question...but the quadratic is just part of it.

When an elastic spring is stretced beyond its natural unstretched length, a tension exists in the spring creating simple harmonic motion. The magnitude of the tension is given by the experimental law credited to Robert Hooke. Hooke's law states that the tension is directly proportional to the extension of the spring beyond its natural length i.e.,

T = k x

In this expression, T denotes the tension measured in Newtons, and x the extension beyond the natural length of the spring. The constant of proportionality, referred to as the spring constant, describes the stuffness of the spring and is stated in Netwons/metre.

Using Simple Harmonic Motion to model 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. This bridge is 71mabove 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 metre.

Participants jump from the bridge, fastened to an elastic rope that is adjusted to half their descent at an appropiate level.

The rope is specially designed and its spring constant is known from specifications. For the purposesof this 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 attacted 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.

4. Aug 23, 2006

### uart

1. You appear to have assumed that m=75kg which is not what you should be doing by my reading of the question. I believe that your answer should be in the form of an expression involving both m and L.

2. If you do assume that m=75 kg then you do get a quadratic somewhat similar to the one you posted. There are some important errors however. I think you'll find that the coefficient "a" should be 1/2 instead of 37.5, the coefficent "b" should be -L and the coefficent "c" should be -L^2. Note the negative signs for the b and c coefficients!

Why dont you show us your working?

PS. You get the quadratic equation by equating the decrease in gravitational potential energy to the increase in the elastic stored PE of the rope. As in 1/2 k x^2 = m g (x+L), where x is the extension of the rope beyond it's unstretched length and k*L is a constant that can be determined from the given data.

Last edited: Aug 23, 2006