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

  1. Jun 14, 2016 #1
    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?
     
  2. jcsd
  3. Jun 14, 2016 #2
    Do you know any of the standard kinematics equations for a constant acceleration?
     
  4. Jun 14, 2016 #3
    K = 1/2mv2?
     
  5. Jun 14, 2016 #4

    PeroK

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    That's not a good answer!
     
  6. Jun 14, 2016 #5
    Sorry for my ignorance.
     
  7. Jun 14, 2016 #6

    PeroK

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    Have you just picked up this formula from Wikipedia without any prior knowledge or are you learning some physics/kinematics?
     
  8. Jun 14, 2016 #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.
     
  9. Jun 14, 2016 #8

    PeroK

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    You've never seen the equation ##s = \frac{1}{2}at^2## then?
     
  10. Jun 14, 2016 #9
    no
     
  11. Jun 14, 2016 #10
    Is that what it's derived from?
     
  12. Jun 14, 2016 #11

    PeroK

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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]
     
    Last edited by a moderator: May 8, 2017
  13. Jun 14, 2016 #12
    Thank You
     
    Last edited by a moderator: May 8, 2017
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