Why is the velocity of a jump on Mars the same as on Earth?

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Homework Help Overview

The discussion revolves around a homework question regarding the height a person can jump on Mars compared to Earth, specifically questioning the assumption that the jump velocity is the same on both planets. The subject area includes physics concepts related to kinematics and biomechanics.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • The original poster attempts to understand why the jump velocity is assumed to be the same on Mars and Earth, questioning the implications of muscle strength and gravitational differences. Some participants discuss the simplifications made in homework models, while others explore the definition of jump height and the work done by leg muscles.

Discussion Status

The discussion is ongoing, with participants providing insights into the assumptions behind the problem. Some guidance has been offered regarding the simplifications in modeling jump dynamics, and there is an exploration of how jump height is defined in relation to gravitational forces.

Contextual Notes

Participants are navigating the constraints of homework rules, which may limit the depth of exploration into the physics involved. There are also discussions about the nonlinear responses of muscles and the implications of different gravitational forces on jump dynamics.

HAF
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Homework Statement


Good evening,
I have a homework where the question is "How high will person jump on Mars?"

Homework Equations


v2 = 2.amars.h1
V2 = 2.g.h2

The Attempt at a Solution


I know everything to solve it. I can solve it by h1 / h2 . In this created equation the only unknown is h1 so it's all cool. But why can I assume that v1 and V1 are the same? Why jump velocity on Mars is the same as on Earth? Aren't there any changes? For example lower strength of muscles due to different weight(force)? I know that maybe my question is completely stupid and I'm very sorry about it. I'm happy that I've atleast tried to find it out but I think that I can't figure out why the velocities are the same without your help.
 
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HAF said:
But why can I assume that v1 and V1 are the same?
Because it is homework and one is expected to use models that are simplified to the point of being incorrect.
 
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jbriggs444 said:
Because it is homework and one is expected to use models that are simplified to the point of being incorrect.
Thank you very much sir. Now everything makes sense to me. I really appreciate it!
 
HAF said:
Why jump velocity on Mars is the same as on Earth?
The assumption is that the work the legs can do is the same. There is also the issue of how the height of jump is defined, but vertical displacement of mass centre (from the crouch position) is the biophysical standard. So it is not really about take off speed. How is that defined? At the moment the feet leave the ground, the Martian jumper would have a greater speed, having done less work against gravity already.

Of course, it is not really true that the same work will be done. Muscles have a couple of limits - max force and max contraction rate - with a nonlinear response along the way.
 
haruspex said:
The assumption is that the work the legs can do is the same. There is also the issue of how the height of jump is defined, but vertical displacement of mass centre (from the crouch position) is the biophysical standard. So it is not really about take off speed. How is that defined? At the moment the feet leave the ground, the Martian jumper would have a greater speed, having done less work against gravity already.

Of course, it is not really true that the same work will be done. Muscles have a couple of limits - max force and max contraction rate - with a nonlinear response along the way.
Aha. So for correctness should I solve it from equation m.g.h(1) = m.a.h(2) right?
 
haruspex said:
There is also the issue of how the height of jump is defined, but vertical displacement of mass centre (from the crouch position) is the biophysical standard.

Or, you could refer to the rules of the Olympic Standing Jump, from 1912:

 
HAF said:
Aha. So for correctness should I solve it from equation m.g.h(1) = m.a.h(2) right?
That's how I would approach it.
 

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