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Galileos law of free fall

  1. Nov 23, 2009 #1
    I'm reading a maths book called Thomas Calculus and in their method for getting the average speed of an object when only the height its dropped from is known is this formula here which they call Galileos law:
    y = 16t^2
    y being the distance travelled after time. What I don't get is where they get the 16 from. All they say about it is "where 16 is the constant of proportionality". Where did they get this constant of proportionality from and does this 16 apply to all falling object scenarios?
  2. jcsd
  3. Nov 23, 2009 #2


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    The general form would be y = .5at^2, where a is the acceleration. The acceleration due to gravity (free fall) is 32 feet/sec^2. So the formula you have been given is for time in seconds and distance in feet, when dropping an object from a height (not too large, since "a" will depend on height) above the surface of the earth.
  4. Nov 23, 2009 #3
    Thanks that explains where they got 16 but why .5?
  5. Nov 23, 2009 #4


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    The .5 comes from solving the Differential equation which describes a free falling body.
  6. Nov 23, 2009 #5


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    And if you're reading a calculus problem, the time derivative of that position would give you [tex]y' = a*t = v[/tex] which is obviously your speed at any given time, t, given a constant acceleration, a, if begun at rest.
  7. Nov 23, 2009 #6
    Ah right. Thanks.
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