Solve Trampoline Question: Jump from 10m & 20m Height

In summary, the problem involves a person jumping from a platform onto a trampoline and calculating how much the trampoline will stretch based on the principles of conservation of energy and the behavior of a simple elastic spring. The correct equation to use is mgh1 = 0.5kx2^2, where m represents the person's mass, g is the acceleration due to gravity, h1 is the height from which the person jumps, and x2 is the amount of stretch in the trampoline. The incorrect use of Fg/x to calculate the spring constant is due to not accounting for the acceleration of the person as they land on the trampoline.
  • #1
vpv
23
0

Homework Statement



A 60 kg person jumps from a platform onto a trampoline 10 m below, stretching it 1.0 m from its initial position. Assumin that the trampoline behaves like a simple elastic spring, how much will it stretch if the same person jumps from a height of 20 m?


Homework Equations


Law of Conservation of Energy
Ek = 0.5mv^2
Eg = mgh
Ee = 0.5kx^2
Fx = kx

The Attempt at a Solution


Really didn't understand the question at all. I figured since the guy is jumping of the platform, velocity at that instance would be 0. If the guy landed on the trampoline, his velocity would be 0 (but i don't know about this). The equillibrium of spring is 0 when the guy jumps of platform so there is no elastic potential. I am guessing there is some gravitational potential as the platform is above the trampoline. But I am consistently getting the wrong answer. The answer is 1.4 m. How would you get that?

Thanks, I appreciate the help.
 
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  • #2
Can you start by writing out an equation representing the principle of conservation of energy?
 
  • #3
Ek1 + Eg1 + Ee1 = Ek2 +Eg2 +Ee2
0.5mv1^2 + mgh1 + 0.5kx1^2 +FaD = 0.5mv2^2 + mgh2 + 0.5kx2^2 +FfD
 
  • #4
vpv said:
Ek1 + Eg1 + Ee1 = Ek2 +Eg2 +Ee2
0.5mv1^2 + mgh1 + 0.5kx1^2 +FaD = 0.5mv2^2 + mgh2 + 0.5kx2^2 +FfD
I think you're over-complicating things a little here. If we ignore air resistance, then there are no dissipative forces present and therefore the total energy of the system must remain constant.

So, what type of energy does the system posses before the person jumps and what type of energy does the system posses after the person has landed on the trampoline?
 
  • #5
Hootenanny said:
I think you're over-complicating things a little here. If we ignore air resistance, then there are no dissipative forces present and therefore the total energy of the system must remain constant.

So, what type of energy does the system posses before the person jumps and what type of energy does the system posses after the person has landed on the trampoline?

Um before the person jumps, he possesses gravitational potential, no? and after he lands, the spring possesses elastic potential.
I think.
 
  • #6
vpv said:
Um before the person jumps, he possesses gravitational potential, no? and after he lands, the spring possesses elastic potential.
I think.
Correct :approve:. So, can you put what you just said in equation form?
 
  • #7
mgh1 = 0.5kx2^2

but how do i find K?
I tried Fg/x which is 588 N/m. I don't know if that is correct tho Caz using that K value, I get a final answer of 6.3 m which is not even close to the answer at the back. Maybe the answer at the back is wrong?
 
  • #8
vpv said:
mgh1 = 0.5kx2^2?
That would be correct. Can you now apply this equation to the two cases:

(1) Where the person jumps from 10m
(2) Where the person jumps from 20m
 
  • #9
Hootenanny said:
That would be correct. Can you now apply this equation to the two cases:

(1) Where the person jumps from 10m
(2) Where the person jumps from 20m

Two words.

You're a Genius.
Wait that's three.

But I want to know why I can't find the k value the way I did. I used Fg/x to find K. Can you tell me what I did wrong?
 
  • #10
vpv said:
Two words.

You're a Genius.
Wait that's three.
:blushing:
vpv said:
But I want to know why I can't find the k value the way I did. I used Fg/x to find K. Can you tell me what I did wrong?
Note that the force exerted by the spring not only opposes the weight of the trampolinist, but must also account for the acceleration of the trampolinist.
 
  • #11
AcceleratioN?? How does he accelerate? Do you mean that he decelerates as he lands on the trampoline and his V = 0?
 
  • #12
vpv said:
AcceleratioN?? How does he accelerate? Do you mean that he decelerates as he lands on the trampoline and his V = 0?
Yes, acceleration is simply a change in velocity - when the speed is increasing or decreasing is irrelevant. Generally in kinematics, one simply talks about acceleration rather than deceleration as things can get a little confusing.
 

1. How do you calculate the velocity needed to jump from a 10m and 20m height on a trampoline?

The velocity needed to jump from a certain height on a trampoline can be calculated using the equation v = √(2gh), where v is the velocity, g is the acceleration due to gravity (9.8 m/s^2), and h is the height of the jump. Plugging in the values for 10m and 20m, we get velocities of approximately 14 m/s and 20 m/s, respectively.

2. Are there any safety precautions that need to be taken when attempting to jump from a 10m and 20m height on a trampoline?

Yes, there are several safety precautions that should be taken when attempting to jump from a high height on a trampoline. These include ensuring that the trampoline is properly set up and secured, using a spotter or safety net, and practicing proper jumping techniques to reduce the risk of injury.

3. What is the maximum height that can be safely jumped from on a trampoline?

The maximum safe jumping height on a trampoline depends on several factors such as the weight and skill level of the jumper, the quality and condition of the trampoline, and the presence of safety measures. Generally, it is recommended to not jump from a height higher than twice the height of the trampoline.

4. Can jumping from a high height on a trampoline cause injury?

Yes, jumping from a high height on a trampoline can cause injury if proper safety precautions are not taken or if the jumper does not have the necessary skill and control. Common injuries from high jumps on trampolines include sprains, strains, and fractures.

5. How does the elasticity of the trampoline affect the height of the jump?

The elasticity of the trampoline, or its ability to stretch and bounce back, plays a crucial role in the height of the jump. A more elastic trampoline will provide a greater upward force, allowing for a higher jump. However, it is important to note that the elasticity should not be too high as it can also increase the risk of injury.

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