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

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?

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

K = 1/2mv2?

PeroK
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K = 1/2mv2?

Sorry for my ignorance.

PeroK
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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?

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
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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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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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