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Terminal velocity

  1. Mar 18, 2017 #1
    Hello,

    I am finishing my IB internal assessment in physics. I have thrown four balls (with different masses, 400,450,475,500 grams) from a height, which is approximately 23 meters. My teacher told me to set up a graph which showed mass vs. vt^2.

    He said that the inverse of M= ((1/2)(p)(C_d)(A) / (g)) x v^2 is the slope. I have problems making anything of this. Can someone please explain how to get up with the equation above?

    Thank you very much!
     
  2. jcsd
  3. Mar 18, 2017 #2
    The purpose of the investigation was to see how terminal velocity acts on objects with different weight.
     
  4. Mar 18, 2017 #3

    PeterDonis

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    Moderator's note: moved to homework forum.
     
  5. Mar 18, 2017 #4

    PeterDonis

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    What is the formula for terminal velocity?
     
  6. Mar 18, 2017 #5
    v_terminal = (Sqrt(2mg)/(C*p_air*A))
     
  7. Mar 18, 2017 #6

    haruspex

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    That is not the slope. The slope is either the ((1/2)(p)(C_d)(A) / (g)) part or its inverse. Which of those depends on how you assign mass and v2 to the x and y axes.
     
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