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Calculating Gravitational Acceleration

  1. Sep 27, 2010 #1
    I'm not 100% if I had worded the title correctly, but I was wondering how to calculate the acceleration of a ball dropping at height.

    I have it's weight, time and height.
    0.45kg
    67 milliseconds
    3m height
     
  2. jcsd
  3. Sep 28, 2010 #2
    [tex] d=(.5)gt^2 [/tex]


    d=distance
    t=time
    g= acceleration due to gravity
    and i'm sure you know that mass doesn't matter
     
  4. Sep 28, 2010 #3

    HallsofIvy

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    Science Advisor

    The original equation was about calculating the acceleration. For an object, like a ball, dropped at the earth's surface, the acceleration is that constant, g, approximately 9.81 m/s2, in cragar's post.
     
  5. Sep 28, 2010 #4
    That's what I was thinking. But it dropped from a height of 3m in 67 milliseconds. I did the above equation:

    d = (0.5)gt^2
    g = 9.8
    t = 0.67

    Therefore,

    0.5 x 9.8 x 0.67 x 0.67 should equate to 3 - or am I mistaken and/or made some errors?
     
  6. Sep 29, 2010 #5
    67ms = 0.067s not 0.67
    And how did you measure the time? You should repeat the experiment several times and see how big is the error.
     
  7. Sep 29, 2010 #6
    It was recorded on a Digital Camera and I viewed it through Windows Movie Maker. I watched it again several times it's definitely within the 60 - 70 millisecond range.
     
  8. Sep 30, 2010 #7

    cjl

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    You mean 670-700 millisecond range.


    1 millisecond = 1/1000 of a second. There's no way that it fell 3 meters in 67 milliseconds, at least not on earth.
     
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