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

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The discussion focuses on 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 emphasize the importance of understanding the formulas rather than just plugging in numbers for an answer. Key equations discussed include v = u + at and x = ut + 1/2 at², with the initial velocity (u) being zero. One user initially calculated the time as 0.83 seconds, but another corrected it to 1.08 seconds after proper calculations. Understanding the context and values in the equations is crucial for solving the problem accurately.
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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
 
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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 V^2 = U^2 + 2as

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

x = ut + \frac{1}{2} at^2

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.
 
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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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