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t = √(2d/g).

Why is the distance doubled in this equation/ how is this derived?

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- Thread starter Jude Caird
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- #1

- 7

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t = √(2d/g).

Why is the distance doubled in this equation/ how is this derived?

- #2

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Do you know any of the standard kinematics equations for a constant acceleration?

- #3

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K = 1/2mv^{2}?

- #4

- 16,688

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K = 1/2mv^{2}?

That's not a good answer!

- #5

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Sorry for my ignorance.

- #6

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Sorry for my ignorance.

Have you just picked up this formula from Wikipedia without any prior knowledge or are you learning some physics/kinematics?

- #7

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- #8

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You've never seen the equation ##s = \frac{1}{2}at^2## then?

- #9

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noYou've never seen the equation ##s = \frac{1}{2}at^2## then?

- #10

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Is that what it's derived from?You've never seen the equation ##s = \frac{1}{2}at^2## then?

- #11

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Is that what it's derived from?

Yes, you could google for SUVAT or "kinematics". Or, try this:

http://www.examsolutions.net/maths-revision/mechanics/kinematics/suvat/derivation/tutorial-1.php [Broken]

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- #12

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Thank YouYes, you could google for SUVAT or "kinematics". Or, try this:

http://www.examsolutions.net/maths-revision/mechanics/kinematics/suvat/derivation/tutorial-1.php [Broken]

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