How to find the height where two same marbles have the same velocity?

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Homework Help Overview

The discussion revolves around determining the height at which two identical marbles, dropped from different gravitational environments (Earth and Moon), attain the same velocity. The problem involves kinematic equations and considerations of gravitational acceleration.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants explore the interpretation of the problem, questioning whether the velocities are equal upon release or upon impact. There is discussion about the implications of different gravitational accelerations and the absence of air resistance on the Moon.

Discussion Status

Some participants have provided insights into the concept of terminal velocity and its relevance to the problem. There is an ongoing exploration of how to relate the velocities of the two marbles and the heights from which they are dropped, with no explicit consensus reached yet.

Contextual Notes

The problem lacks specific numerical values for height or velocity, which may influence the approach to finding a solution. The absence of drag or air resistance on the Moon is also noted as a significant factor in the discussion.

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Homework Statement



How do you find the height where two same marbles have the same velocity when one marble is dropped from the Earths atmosphere (g=9.8) and the other is dropped from the Moons atmosphere (g=1.6)? Both are dropped from the same height

Homework Equations



kinematic equations

The Attempt at a Solution



I have absolutely no idea where to start...
 
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Is that the question exactly?
 
yes.
 
mwahx3 said:
yes.

Are they saying the velocity is the same as they hit the ground?

That's a strange question... because if they are dropped from exactly the same height, then when they hit the ground they'll both always have different velocities because the g's are different... unless the height they are dropped from is 0...
 
yea that was what i was wondering...no it doesn't say that.
maybe the velocity is the same right when they are released?
the only possible reason i can think of...
 
mwahx3 said:
yea that was what i was wondering...no it doesn't say that.
maybe the velocity is the same right when they are released?
the only possible reason i can think of...

Do they give any other numbers... like the height at which they're released?
 
no they dont. they just say at height 'H'.
 
mwahx3 said:
no they dont. they just say at height 'H'.

Hmmm... that gives the impression that the answer can be given in terms of H... do they also say velocity V or anything like that?
 
no, since they ask for the velocity in one part of the problem...
 
  • #10
mwahx3 said:
no, since they ask for the velocity in one part of the problem...

Do they give drag or air resistance for this question?
 
  • #11
F=-kv

and k= 0.1854
 
  • #12
mwahx3 said:
F=-kv

and k= 0.1854

Ah... ok, now I think I understand... a marble is dropped from a height on earth... due to air resistance it reaches its terminal velocity before hitting the ground... when it hits its terminal velocity, the net force on it is 0... so its acceleration is 0 at that point... and it stays at that terminal velocity...

On the moon, there's no atmosphere... so there's no air resistance to think about.

So, first find the terminal velocity of the marble on earth... then find the height that the marble needs to be dropped from on the moon, so that it reaches that velocity just as it hits the ground.
 
  • #13
ohh okay thank you. I think I got the terminal velocity...but I'm having trouble finding the height the marble needs to be dropped from the moon. What velocity should I use?
 
  • #14
mwahx3 said:
ohh okay thank you. I think I got the terminal velocity...but I'm having trouble finding the height the marble needs to be dropped from the moon. What velocity should I use?

The terminal velocity that you get for the Earth marble.
 
  • #15
ahhh okay! thank you! would that just be my answer?
 
  • #16
mwahx3 said:
ahhh okay! thank you! would that just be my answer?

Yeah, the terminal velocity would be your answer for that part.
 

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