Unknown unstretched elastic cord length

[gosbun]

A person jumps from a balloon 65m above the ground. He will stop 10m above the ground. A 5m length of the same cord stretches 1.5m with his body weight. He will drop from rest at the point where the top end of a longer section of the cord is attached to the stationary balloon. What length of cord should the jumper use?

Is the spring or elastic constant (mg/1.5)? If it is how do I set up the equation?

I know that mgyi=65m, and mgyf=10m.

 

[Doc Al]

Is the spring or elastic constant (mg/1.5)? If it is how do I set up the equation?

That's the spring constant for a 5m length of cord; the spring constant is inversely proportional to the cord length.

The key to this problem is energy conservation: The initial gravitational PE is transformed to spring potential energy. (Spring PE is given by 1/2kx^2.) Realize that both the maximum spring extension (x) and the spring constant are functions of the cord length: Set up the equation and you can solve for that cord length.

 

[wais]

The length of the elastic cord that the jumper should use can be calculated by setting up the equation for the conservation of energy. The equation is:

(mgh)i = (mgh)f + 0.5kx^2

Where m is the mass of the jumper, g is the acceleration due to gravity, h is the initial height (65m), and x is the unknown length of the elastic cord.

Since the cord is attached to the stationary balloon, the length of the cord will not change during the jump. Therefore, the change in height (h) will be equal to the change in length of the cord (x).

We can also substitute the value for the elastic constant (k), which is equal to mg/1.5, giving us the final equation:

(mgh)i = (mgh)f + 0.5(mg/1.5)(x^2)

Substituting the given values, we get:

(m)(9.8)(65) = (m)(9.8)(10) + 0.5(m)(9.8/1.5)(x^2)

Solving for x, we get x = 11.5m. Therefore, the jumper should use a 11.5m length of elastic cord to stop 10m above the ground.

It is important to note that the given values assume that the elastic constant (k) is equal to mg/1.5. If this is not the case, the equation and the final answer may differ.