# Bungee jumping problem with a twist

• harihrn
In summary, the man will travel at 24.3 m/s when he hits the ramp on the scooter, assuming no air resistance.
harihrn
Warning: Homework template missing.
A bungee jumping gig is about to be performed off a bridge at a height of 100ft. The length of the rope is 20ft. The K constant for the rope is 200n/m and the air resistance is 12N(only vertical). The jumper's mass is 100kg. When the man jumps, the rope snaps right before it starts to stretch, making him land on a scooter on top of a ramp. If the ramp is elevated 5 degrees, and the length of his path down the ramp is 50 ft, how fast will he be going as he runs down the ramp on to the surface of earth? (assume 0 mass for scooter and no friction on ramp)

I attempted the problem thinking that the KE from the fall would convert into the KE during the run on the ramp.

so i did Ke+Fe(15.24) = 1/2mv^2; with fe being sin(5degrees)m*g
Ke = mg(total height-height of ramp) - 12(total height-height of ramp);
12 being the force of friction.
i got 24.3 m/s as the answer.
But when this was attempted on a simulator, it showed a rise in thermal energy, which is essentially work i did not account for. So how would i exactly approach this problem?

harihrn said:
the rope snaps right before it starts to stretch
Do you mean that? If so, the K constant is irrelevant; in fact, the rope irrelevant. Perhaps you mean it snaps at full stretch?

I attempted the problem thinking that the KE from the fall would convert into the KE during the run on the ramp.

That PE to KE conversion is a great method if there are no energy losses - but you have the air resistance in this problem so you have to take that into account, perhaps using forces and acceleration to work out the speed at the end of the fall.

The collision with the scooter is puzzling. In reality, the body would probably deform badly and not survive. Maybe you are expected to assume that the collision force on the body is in one direction only so momentum could be conserved in the perpendicular direction.

Greg Bernhardt
Delphi51 said:
I attempted the problem thinking that the KE from the fall would convert into the KE during the run on the ramp.

That PE to KE conversion is a great method if there are no energy losses - but you have the air resistance in this problem so you have to take that into account, perhaps using forces forces and acceleration to work out the speed at the end of the fall.

The collision with the scooter is puzzling. In reality, the body would probably deform badly and not survive. Maybe you are expected to assume that the collision force on the body is in one direction only so momentum could be conserved in the perpendicular direction.
You didn't answer my question. If it snaps just before it starts to stretch then it has achieved nothing and you can treat it as though the rope never existed. Besides, why would a rope snap when under no tension?
But changing it to "the rope snaps right before it starts to contract", i.e. it snaps at full stretch, doesn't help - you can then ignore the fall and just place the jumper gently on the bike at the top of the slope.
I feel we don't have the correct statement of the problem yet.

yeah it's part of the problem, probably a distraction. it's right before the rope starts to stretch. and also, it is to be assumed that the guy survives the fall

The mass of the scooter isn't given so it appears impossible to work out what their combined speed is just after the collision. Presumably you will have to assume the mass is zero?

Edit: oops I see it says to ignore the mass of the scooter. In which case you can assume they start at the velocity of the man.

I think the approach you've used can be described in this way (as noticed, the rope and its K constant seem to be irrelevant):

The considerations being taken into account are: a change in the ''roughness'' of the floor and a change in the magnitude of the force down the slope. I think your simulator is accounting for a lost of thermal energy because it is truly considering the problem as one of a person falling. Then, what we're just considering as a change in roughness when going from the ##100ft-h##-floor to the ##20ft##-floor, the simulator is considering it as an impact, where the Normal Force is not mgcos(theta) and the downward acceleration is not mgsin(theta) when he first hits the top of the slope of 20ft.

By Newton's Second Law I find the accelerations in each section and then their velocities usign kinematic formulas. The final one at the bottom agrees with that you've found, but I suspect is not the right way to solve the problem but one where an impact is taken into account (I have no clue about how to do it).

Jazz said:
I think the approach you've used can be described in this way (as noticed, the rope and its K constant seem to be irrelevant):

I can't relate your diagram to the given facts. He falls vertically 100ft (yes?) onto a scooter at the top of a 5 degree ramp length 50ft. The 20ft is irrelevant.
Note that the velocity with which he hits the scooter does not become an initial velocity down the slope. The component of it normal to the slope will be lost as thermal energy.

I agree there is no force exerted by the rope. Just gravity and air resistance during the fall. In the absence of any initial velocity, I would treat it as a purely vertical fall. The velocity at the end of the fall could be found with
PE at top - work done against air resistance = KE at bottom
Careful with units - I would convert that 100 ft into meters before writing it down!

haruspex said:
I can't relate your diagram to the given facts. He falls vertically 100ft (yes?) onto a scooter at the top of a 5 degree ramp length 50ft. The 20ft is irrelevant.
Note that the velocity with which he hits the scooter does not become an initial velocity down the slope. The component of it normal to the slope will be lost as thermal energy.
I assumed the 100ft-height is measured from the ground. While falling there is a net acceleration ##a##, which is a little less than that of gravity (because of air resistance, which I assumed constant).

The acceleration has that magnitude until he hits the top of the slope. Here, the last chunk of the 100ft-path is not traveled with the acceleration ##a##. That's why I subtracted h from 100ft.

I confused 20ft with 50ft. The latter is the correct. I will recalculate it.

As far as I understand, Ncos(theta) remove some of the speed that goes into thermal energy, so there is no conservation of energy, but how long does the overall apparent weight act? or how can the distance along which it does work be figured out?

Jazz said:
As far as I understand, Ncos(theta) remove some of the speed that goes into thermal energy,
It's an inelastic impact, so consider impulse, not force. You can just throw away the component of speed normal to the slope and keep the bit parallel to it.

but the momentum isn't transferred to the ramp itself is it? the ramp doesn't move at all

harihrn said:
but the momentum isn't transferred to the ramp itself is it? the ramp doesn't move at all
The momentum normal to the ramp will be transferred to the ramp+earth+rider system, bringing them all to the same velocity in that direction. But for all practical purposes it just disappears.

## 1. What is the "twist" in the bungee jumping problem?

The twist in the bungee jumping problem refers to an additional factor that complicates the traditional calculations for bungee jumping, such as wind resistance, varying elasticity of the bungee cord, or human error.

## 2. How does the "twist" affect the safety of bungee jumping?

The "twist" in the bungee jumping problem can potentially make the activity more dangerous as it introduces an element of uncertainty that cannot be accounted for in traditional safety measures. It is important for bungee jumping operators to thoroughly assess and mitigate any potential risks associated with the "twist".

## 3. Can the "twist" be calculated and accounted for in bungee jumping?

In most cases, the "twist" in the bungee jumping problem can be calculated and accounted for with advanced mathematical models and simulations. However, these calculations can only provide an estimation and cannot account for all potential variables. It is important for bungee jumping operators to regularly review and update their safety protocols based on new information and advancements in technology.

## 4. Are there any regulations or standards for bungee jumping with a "twist"?

Currently, there are no specific regulations or standards for bungee jumping with a "twist". However, bungee jumping operators are required to adhere to general safety regulations and obtain proper permits and licenses from local authorities. It is also recommended for operators to follow industry best practices and stay informed about any advancements or changes in safety protocols.

## 5. What precautions should be taken when bungee jumping with a "twist"?

To ensure maximum safety when bungee jumping with a "twist", it is important for bungee jumping operators to conduct thorough risk assessments, regularly maintain and inspect equipment, and provide proper training to staff. It is also recommended for individuals to carefully consider their physical and mental readiness before participating in bungee jumping and to follow all safety instructions provided by the operator.

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