# Time taken for free falling objects to travel a set distance

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## Main Question or Discussion Point

The time taken for a free-falling object to fall a set distance (on earth) is, according to Wikipedia, described by the equation:

t = √(2d/g).

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

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

K = 1/2mv2?

PeroK
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Gold Member
K = 1/2mv2?

Sorry for my ignorance.

PeroK
Homework Helper
Gold Member
Sorry for my ignorance.
Have you just picked up this formula from Wikipedia without any prior knowledge or are you learning some physics/kinematics?

I am interested in orbital mechanics etc. I have a suitable understanding of physics and kinematics and the formulae associated with gravity. I tried to derive the formula from my own knowledge, but failed.

PeroK
Homework Helper
Gold Member
I am interested in orbital mechanics etc. I have a suitable understanding of physics and kinematics and the formulae associated with gravity. I tried to derive the formula from my own knowledge, but failed.
You've never seen the equation $s = \frac{1}{2}at^2$ then?

You've never seen the equation $s = \frac{1}{2}at^2$ then?
no

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

PeroK
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Gold Member
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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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]
Thank You

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