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Old Jan 8th 2009, 10:42 AM   #1
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speed of a feather falling.

A feather is placed in a 2 m long vacuum tube that has all the air removed from it. If you flip the tube upside-down how long will it take for the feather to fall to the bottom of the tube?
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Old Jan 8th 2009, 09:54 PM   #2
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Originally Posted by ariol View Post
A feather is placed in a 2 m long vacuum tube that has all the air removed from it. If you flip the tube upside-down how long will it take for the feather to fall to the bottom of the tube?
Taking downward as positive
$\displaystyle s=ut+gt^2/2$
$\displaystyle 2=0+10t^2$
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Old Jan 9th 2009, 06:17 AM   #3
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Since the air has been removed, there are no opposing forces acting on the feather. The feather moves in a straight line.

So, we can use one of the equations of linear motion. We know $\displaystyle s = 2\ m $. We know that the initial velocity of the feather is $\displaystyle 0$, i.e. $\displaystyle u = 0 $. $\displaystyle a = g = 10\ \ or \ \ 9.81 ms^{-2}$.

$\displaystyle s = ut + \frac {1}{2} at^{2} $

$\displaystyle 2 = 0 \times t + \frac {1}{2} \times 9.81 \times t^{2} $

$\displaystyle 2 = \frac {1}{2} \times 9.81 \times t^{2}$

$\displaystyle t = \sqrt {\frac {2}{\frac {1}{2} \times 9.81}} $

$\displaystyle t = 0.639 s $ (correct to 3 sig. fig.)

With $\displaystyle g = 10$, you get $\displaystyle t = 0.632 s$.

I hope that helps.

ILoveMaths07.

Last edited by ILoveMaths07; Jan 9th 2009 at 06:20 AM.
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