Time taken for free falling objects to travel a set distance

  • #1

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?
 

Answers and Replies

  • #2
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Do you know any of the standard kinematics equations for a constant acceleration?
 
  • #3
K = 1/2mv2?
 
  • #4
PeroK
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  • #5
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
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.
 
  • #8
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?
 
  • #9
You've never seen the equation ##s = \frac{1}{2}at^2## then?
no
 
  • #10
You've never seen the equation ##s = \frac{1}{2}at^2## then?
Is that what it's derived from?
 
  • #11
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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  • #12
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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