Bungy Jumping in realtion to Hooke's law

[a14652289]

A person is going bungy jumping. there is 2 kinds. Wet and Dry. in wet the person is submerged to a depth of about 1m. in dry, the person fall ends just above the surface of water. In one case, the bridge os 43m above the river and in the other case its 71m above the river.

Participants jump fromt he bridge, fastened to an elastic rope which is adjuseted to halt their descent at the appropriate level.

The rope is specially designed and its modulis of elasticity is known from specifications. For the purpose of this problem, we will assume th rope is streched to twice its natural length bu a person of mass 75kg hanging at the rest from the free end. It is necessary to adjust the length of the rope in terms of th weight of the jumper.

1) For a preson of mass m kg, calculate the depth of to which the person would fall if attached to a rope of the type described above, with length (l) metres.

2) FOr the 71m jump, aompare (l) as a function of m for the following twwo jumps:
A dry jump, where the descent is halted 1m above the water and a wet jump, where the descent is halted below the surface.


equations that maybe of use:

T = kx (T=tension, k= is know as the stiffness or spring constant, x= extension beyond natural length)

T = (lander / l)x (lander isthe modulus of elasticity and l = natural length)

Force = mass x accerleration

Also derivatives and integration is used.

 

[tiny-tim]

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Hi a14652289! Welcome to PF!

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[tomcue]

I would first like to clarify the concept of Hooke's law in relation to bungy jumping. Hooke's law states that the force applied to an elastic object is directly proportional to the amount of extension or compression of the object. In the case of bungy jumping, the elastic rope acts as the object and the force applied is the person's weight.

Now, let's consider the two types of bungy jumping mentioned - wet and dry. In both cases, the person is attached to an elastic rope which is adjusted to halt their descent at a certain level. The difference lies in the depth of the water the person falls into.

In the wet jump, the person is submerged to a depth of about 1m, while in the dry jump, the person falls just above the surface of the water. This means that in the wet jump, the person experiences a greater resistance force due to the water, compared to the dry jump where there is no resistance from the water.

Using the equations provided, we can calculate the depth to which a person would fall if attached to a rope of length (l) meters for any given mass (m) kg. We would need to consider the force of gravity (mg) acting on the person, as well as the tension (T) in the rope, which is equal to the force of gravity.

1) For a person of mass m kg, the depth to which they would fall can be calculated using the equation T = kx. We know that the tension in the rope is equal to the force of gravity, so we can substitute T with mg. Rearranging the equation, we get x = mg/k. This gives us the extension of the rope, which is equal to the depth of the fall. Therefore, the depth of the fall for a person of mass m kg would be x meters.

2) For the 71m jump, we can compare the length of the rope (l) as a function of mass (m) for both the wet and dry jumps. Using the equation T = (lander/l)x, we can rearrange it to get l = (lander/mg)x. This gives us the length of the rope required for a person of mass m kg to jump from a height of 71m and come to a stop at a depth of x meters. We can then compare the values of l for both the wet and dry jumps to see the difference in