- #1

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Why is this?

Please involve calculations if possible. I would much appreciate any form of help.

- Thread starter narbij
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- #1

- 7

- 0

Why is this?

Please involve calculations if possible. I would much appreciate any form of help.

- #2

- 489

- 0

When you introduce fluid resistance, ie air resistance, you get two things happening. Firstly, obviously, you get the larger (surface area wise) objects being slowed more and secondly you get terminal velocity limiting the speed of the object.

Calculation wise newton tells us:

[tex] {v^2} = {u^2} + 2as[/tex]

Thus if initial velocity, displacement and acceleration are equal for all objects, regardless of mass, final velocity will be the same as well.

In terms of air resistance, the equations differ for different shapes. For a sphere of diameter D the initial equation is the following:

[tex]F(v) = -c1v - c2v *|v|[/tex] (sorry still gettin used to tex, c1 and c2 are constants )

These constants are experimentally determined and have the following values:

c1 = [tex]1.55 x {10^(-4)} * {D^2}[/tex]

c2 = [tex]0.22 {D^2}[/tex]

Thus usually c2 is used as the c1 value is small enough to drop out. By using c2 we realise we are dealing with a quadratic situation.

hope this makes sense

-G

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