Dynamics - Bungee Jumping - Grade 12 Maths C
Clash Royale CLAN TAG#URR8PPP
up vote
2
down vote
favorite
The full question is as follows:
New Zealand is the home of bungee jumping. One of the major jumps is located on a bridge over the Shotover River near Queenstown.
In this case, the bridge is 71 m above the river.
Participants jump from the bridge, fastened to an elastic rope that is adjusted to halt their descent at an appropriate level.
The rope is specially designed and its spring constant is known from specifications. For the purposes of the 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.
- For a person of mass $m$ kg, calculate the depth to which a person would fall if attached 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. Discuss the assumptions made in this calculation.
- Find the length of rope needed for a wet jump, where the descent would end 1 m below the surface of the water. Find the speed of entry to the water.
- At present, the model does not include air resistance. Discuss the changes which would have to be made to the model to include air resistance, which is proportional to the velocity of the jumper. Discuss the difficulties involved with the mathematics of this model.
- Read the newspaper article entitled “Bungee jumping requires leap of faith†(available from the Bungee.com website, www.bungee.com/bzapp/press/lj.html). Use
mathematics to support or refute the journalist’s comments.
Please correct me if I‘m wrong, but so far, I have used $k=75g/l$ to find that $textdepth = l(1 + sqrt2m/75)$.
I then applied this equation to the first part of question 2. However, I am stuck as to how I find the velocity of the jumper when they enter the water. My current attempt is to use $textangular frequency = sqrtk/m$ and then to use this in the equation $v=d·w$. In this case, $d=1$, therefore giving me a velocity of about 0.62 m/s. This answer seems right, but I’m not sure if my working is correct...
For question 3, the only way I know how to include air resistance is $F_textnet = F_textair + kx -mg$, but I am unsure how to use this or where to go from this point.
physics
edited Aug 17 '17 at 14:33
Wrzlprmft
2,86411233
asked Aug 17 '17 at 6:50
greenie
344
add a comment |Â
up vote
2
down vote
favorite
The full question is as follows:
New Zealand is the home of bungee jumping. One of the major jumps is located on a bridge over the Shotover River near Queenstown.
In this case, the bridge is 71 m above the river.
Participants jump from the bridge, fastened to an elastic rope that is adjusted to halt their descent at an appropriate level.
The rope is specially designed and its spring constant is known from specifications. For the purposes of the 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.
- For a person of mass $m$ kg, calculate the depth to which a person would fall if attached 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. Discuss the assumptions made in this calculation.
- Find the length of rope needed for a wet jump, where the descent would end 1 m below the surface of the water. Find the speed of entry to the water.
- At present, the model does not include air resistance. Discuss the changes which would have to be made to the model to include air resistance, which is proportional to the velocity of the jumper. Discuss the difficulties involved with the mathematics of this model.
- Read the newspaper article entitled “Bungee jumping requires leap of faith†(available from the Bungee.com website, www.bungee.com/bzapp/press/lj.html). Use
mathematics to support or refute the journalist’s comments.
Please correct me if I‘m wrong, but so far, I have used $k=75g/l$ to find that $textdepth = l(1 + sqrt2m/75)$.
I then applied this equation to the first part of question 2. However, I am stuck as to how I find the velocity of the jumper when they enter the water. My current attempt is to use $textangular frequency = sqrtk/m$ and then to use this in the equation $v=d·w$. In this case, $d=1$, therefore giving me a velocity of about 0.62 m/s. This answer seems right, but I’m not sure if my working is correct...
For question 3, the only way I know how to include air resistance is $F_textnet = F_textair + kx -mg$, but I am unsure how to use this or where to go from this point.
physics
edited Aug 17 '17 at 14:33
Wrzlprmft
2,86411233
asked Aug 17 '17 at 6:50
greenie
344
How did you get that formula for depth? I got a different result.
– Aretino
Aug 17 '17 at 14:55
For 3 it just asks for discussion, not a mathematical solution. You would just say that air resistance will slow the jumper down and reduce the depth. You are correct in how you add in the air resistance but maybe should show that it leads to a differential equation because you can't assume energy conservation any more. It turns out to be a hard one but that is probably beyond the scope of the question.
– Ross Millikan
Aug 17 '17 at 15:20
Maybe of interest real-world-physics-problems.com/physics-of-bungee-jumping.html
– cgiovanardi
Aug 18 '17 at 1:33
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
The full question is as follows:
New Zealand is the home of bungee jumping. One of the major jumps is located on a bridge over the Shotover River near Queenstown.
In this case, the bridge is 71 m above the river.
Participants jump from the bridge, fastened to an elastic rope that is adjusted to halt their descent at an appropriate level.
The rope is specially designed and its spring constant is known from specifications. For the purposes of the 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.
- For a person of mass $m$ kg, calculate the depth to which a person would fall if attached 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. Discuss the assumptions made in this calculation.
- Find the length of rope needed for a wet jump, where the descent would end 1 m below the surface of the water. Find the speed of entry to the water.
- At present, the model does not include air resistance. Discuss the changes which would have to be made to the model to include air resistance, which is proportional to the velocity of the jumper. Discuss the difficulties involved with the mathematics of this model.
- Read the newspaper article entitled “Bungee jumping requires leap of faith†(available from the Bungee.com website, www.bungee.com/bzapp/press/lj.html). Use
mathematics to support or refute the journalist’s comments.
Please correct me if I‘m wrong, but so far, I have used $k=75g/l$ to find that $textdepth = l(1 + sqrt2m/75)$.
I then applied this equation to the first part of question 2. However, I am stuck as to how I find the velocity of the jumper when they enter the water. My current attempt is to use $textangular frequency = sqrtk/m$ and then to use this in the equation $v=d·w$. In this case, $d=1$, therefore giving me a velocity of about 0.62 m/s. This answer seems right, but I’m not sure if my working is correct...
For question 3, the only way I know how to include air resistance is $F_textnet = F_textair + kx -mg$, but I am unsure how to use this or where to go from this point.
physics
edited Aug 17 '17 at 14:33
Wrzlprmft
2,86411233
asked Aug 17 '17 at 6:50
greenie
344
The full question is as follows:
New Zealand is the home of bungee jumping. One of the major jumps is located on a bridge over the Shotover River near Queenstown.
In this case, the bridge is 71 m above the river.
Participants jump from the bridge, fastened to an elastic rope that is adjusted to halt their descent at an appropriate level.
The rope is specially designed and its spring constant is known from specifications. For the purposes of the 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.
- For a person of mass $m$ kg, calculate the depth to which a person would fall if attached 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. Discuss the assumptions made in this calculation.
- Find the length of rope needed for a wet jump, where the descent would end 1 m below the surface of the water. Find the speed of entry to the water.
- At present, the model does not include air resistance. Discuss the changes which would have to be made to the model to include air resistance, which is proportional to the velocity of the jumper. Discuss the difficulties involved with the mathematics of this model.
- Read the newspaper article entitled “Bungee jumping requires leap of faith†(available from the Bungee.com website, www.bungee.com/bzapp/press/lj.html). Use
mathematics to support or refute the journalist’s comments.
Please correct me if I‘m wrong, but so far, I have used $k=75g/l$ to find that $textdepth = l(1 + sqrt2m/75)$.
I then applied this equation to the first part of question 2. However, I am stuck as to how I find the velocity of the jumper when they enter the water. My current attempt is to use $textangular frequency = sqrtk/m$ and then to use this in the equation $v=d·w$. In this case, $d=1$, therefore giving me a velocity of about 0.62 m/s. This answer seems right, but I’m not sure if my working is correct...
For question 3, the only way I know how to include air resistance is $F_textnet = F_textair + kx -mg$, but I am unsure how to use this or where to go from this point.
physics
physics
edited Aug 17 '17 at 14:33
Wrzlprmft
2,86411233
asked Aug 17 '17 at 6:50
greenie
344
edited Aug 17 '17 at 14:33
Wrzlprmft
2,86411233
asked Aug 17 '17 at 6:50
greenie
344
edited Aug 17 '17 at 14:33
Wrzlprmft
2,86411233
edited Aug 17 '17 at 14:33
Wrzlprmft
2,86411233
edited Aug 17 '17 at 14:33
Wrzlprmft
2,86411233
2,86411233
asked Aug 17 '17 at 6:50
greenie
344
asked Aug 17 '17 at 6:50
greenie
344
asked Aug 17 '17 at 6:50
greenie
344
344
How did you get that formula for depth? I got a different result.
– Aretino
Aug 17 '17 at 14:55
For 3 it just asks for discussion, not a mathematical solution. You would just say that air resistance will slow the jumper down and reduce the depth. You are correct in how you add in the air resistance but maybe should show that it leads to a differential equation because you can't assume energy conservation any more. It turns out to be a hard one but that is probably beyond the scope of the question.
– Ross Millikan
Aug 17 '17 at 15:20
Maybe of interest real-world-physics-problems.com/physics-of-bungee-jumping.html
– cgiovanardi
Aug 18 '17 at 1:33
add a comment |Â
How did you get that formula for depth? I got a different result.
– Aretino
Aug 17 '17 at 14:55
For 3 it just asks for discussion, not a mathematical solution. You would just say that air resistance will slow the jumper down and reduce the depth. You are correct in how you add in the air resistance but maybe should show that it leads to a differential equation because you can't assume energy conservation any more. It turns out to be a hard one but that is probably beyond the scope of the question.
– Ross Millikan
Aug 17 '17 at 15:20
Maybe of interest real-world-physics-problems.com/physics-of-bungee-jumping.html
– cgiovanardi
Aug 18 '17 at 1:33
How did you get that formula for depth? I got a different result.
– Aretino
Aug 17 '17 at 14:55
How did you get that formula for depth? I got a different result.
– Aretino
Aug 17 '17 at 14:55
For 3 it just asks for discussion, not a mathematical solution. You would just say that air resistance will slow the jumper down and reduce the depth. You are correct in how you add in the air resistance but maybe should show that it leads to a differential equation because you can't assume energy conservation any more. It turns out to be a hard one but that is probably beyond the scope of the question.
– Ross Millikan
Aug 17 '17 at 15:20
For 3 it just asks for discussion, not a mathematical solution. You would just say that air resistance will slow the jumper down and reduce the depth. You are correct in how you add in the air resistance but maybe should show that it leads to a differential equation because you can't assume energy conservation any more. It turns out to be a hard one but that is probably beyond the scope of the question.
– Ross Millikan
Aug 17 '17 at 15:20
Maybe of interest real-world-physics-problems.com/physics-of-bungee-jumping.html
– cgiovanardi
Aug 18 '17 at 1:33
Maybe of interest real-world-physics-problems.com/physics-of-bungee-jumping.html
– cgiovanardi
Aug 18 '17 at 1:33
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
1
down vote
Some hints:
- Think in terms of energies. In the setup without air resistance, there is no loss of energy. You can calculate the positional energy and stretching energy depending on the depth of the jumper and length of the rope. This suffices to answer questions 1 and 2.
- Forget about oscillations. While there is an oscillation happening here, its only property relevant to the question is amplitude. Calculating anything else from the framework of oscillations is overly complicated.
answered Aug 17 '17 at 15:06
Wrzlprmft
2,86411233
add a comment |Â
protected by Community♦ Sep 8 at 11:43
Thank you for your interest in this question.
Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).
Would you like to answer one of these unanswered questions instead?
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Some hints:
- Think in terms of energies. In the setup without air resistance, there is no loss of energy. You can calculate the positional energy and stretching energy depending on the depth of the jumper and length of the rope. This suffices to answer questions 1 and 2.
- Forget about oscillations. While there is an oscillation happening here, its only property relevant to the question is amplitude. Calculating anything else from the framework of oscillations is overly complicated.
answered Aug 17 '17 at 15:06
Wrzlprmft
2,86411233
add a comment |Â
up vote
1
down vote
Some hints:
- Think in terms of energies. In the setup without air resistance, there is no loss of energy. You can calculate the positional energy and stretching energy depending on the depth of the jumper and length of the rope. This suffices to answer questions 1 and 2.
- Forget about oscillations. While there is an oscillation happening here, its only property relevant to the question is amplitude. Calculating anything else from the framework of oscillations is overly complicated.
answered Aug 17 '17 at 15:06
Wrzlprmft
2,86411233
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Some hints:
- Think in terms of energies. In the setup without air resistance, there is no loss of energy. You can calculate the positional energy and stretching energy depending on the depth of the jumper and length of the rope. This suffices to answer questions 1 and 2.
- Forget about oscillations. While there is an oscillation happening here, its only property relevant to the question is amplitude. Calculating anything else from the framework of oscillations is overly complicated.
answered Aug 17 '17 at 15:06
Wrzlprmft
2,86411233
Some hints:
- Think in terms of energies. In the setup without air resistance, there is no loss of energy. You can calculate the positional energy and stretching energy depending on the depth of the jumper and length of the rope. This suffices to answer questions 1 and 2.
- Forget about oscillations. While there is an oscillation happening here, its only property relevant to the question is amplitude. Calculating anything else from the framework of oscillations is overly complicated.
answered Aug 17 '17 at 15:06
Wrzlprmft
2,86411233
answered Aug 17 '17 at 15:06
Wrzlprmft
2,86411233
answered Aug 17 '17 at 15:06
Wrzlprmft
2,86411233
answered Aug 17 '17 at 15:06
Wrzlprmft
2,86411233
2,86411233
add a comment |Â
add a comment |Â
protected by Community♦ Sep 8 at 11:43
Thank you for your interest in this question.
Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).
Would you like to answer one of these unanswered questions instead?
How did you get that formula for depth? I got a different result.
– Aretino
Aug 17 '17 at 14:55
For 3 it just asks for discussion, not a mathematical solution. You would just say that air resistance will slow the jumper down and reduce the depth. You are correct in how you add in the air resistance but maybe should show that it leads to a differential equation because you can't assume energy conservation any more. It turns out to be a hard one but that is probably beyond the scope of the question.
– Ross Millikan
Aug 17 '17 at 15:20
Maybe of interest real-world-physics-problems.com/physics-of-bungee-jumping.html
– cgiovanardi
Aug 18 '17 at 1:33