How Long Does It Take for a Feather to Fall 1 Meter on the Moon?

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Homework Help Overview

The discussion revolves around calculating the time it takes for a feather to fall 1 meter on the Moon, where the acceleration due to gravity is 1.7 m/s². Participants are exploring the relevant kinematic equations and the implications of the given values.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss which kinematic equations are applicable given the initial conditions and known values. There is a focus on understanding the meaning of the formulas rather than just applying them. Questions about the values provided and what can be inferred from them are raised.

Discussion Status

Some participants have offered guidance on which equations might be useful based on the known variables. There is an ongoing exploration of different approaches to the problem, with no explicit consensus on a single method or answer yet.

Contextual Notes

Participants emphasize the importance of understanding the formulas before applying them, indicating a learning-focused approach. There are also varying answers being proposed, with some participants questioning the correctness of others' calculations.

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On the Moon the acceleration...

On the Moon the acceleration of freefall is 1.7 m s-2. How much time would it take for a featherto fall a distance of 1.0m from rest?

Im not bothered about the answer i just need to know what formula(s) to use
and how.

These are the formulas i was given:

v = u + at
x = ut + 1/2 at^2
v^2 = u^2 +2ax
Average velocity = Displacement / Time
Average Velocity = (v+u)/2

u= initial velocity
a= acceleration
x= displacement
v= final velocity
t= time
 
Last edited:
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bramill said:
On the Moon the acceleration of freefall is 1.7 m s-2. How much time would it take for a featherto fall a distance of 1.0m from rest?

Im not bothered about the answer i just need to know what formula(s) to use
and how.

These are the formulas i was given:

v = u + at
x = ut + 1/2 at^2
v^2 = u^2 +2ax
Average velocity = Displacement / Time
Average Velocity = (v+u)/2

u= initial velocity
a= acceleration
x= displacement
v= final velocity
t= time

But before you can use the formulae, you must understand them, no ?

What do the formulae mean to you ?

What values are you given ? What values can you infer, and what do you need to get ?

There's no point in just pointing out which formula to plug in values into to churn out the answer, that would be like just giving you the answer. You wouldn't learn much from that, I hope you understand.
 
Well you can eliminate the equations that do not give you the information you need.

For example. V = U + at

You know what U is (0 m/s) and you know what a is but you don't know V or t.

So this can't be the equation you need to you use.

What about [tex]V^2 = U^2 + 2as[/tex]

Well, once again we know U and a and for that matter s. So we could use this equation to find V and then plug that into the first equation.

Or Being most sensible you could try

[tex]x = ut + \frac{1}{2} at^2[/tex]

With these questions often the best way to find the answer is to start seeing which ones will give you useful information if you start plugging numbers in.
 
Last edited:
i got an answer of 0.83 s to 2 d.p. can you tell me if this is correct please
 
bramill said:
i got an answer of 0.83 s to 2 d.p. can you tell me if this is correct please

How did you get that (the working) ?
 
the answer is 1.08 i just worked it out properly
 

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