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Isolating t in horizontal motion equation

  1. Jun 17, 2011 #1
    Isolating "t" in horizontal motion equation

    1. The problem statement, all variables and given/known data
    So this is just a practice question from my course and they gave me the answer, the problem I am having is understanding how they got to that answer. I have the formula and everything, just don't know how to isolate "t" and the steps to do so


    -How long does it take for a flowerpot to fall from a balcony that is 18m above the ground?


    2. Relevant equations

    d=½a[itex]\nabla[/itex]t²

    3. The attempt at a solution

    18m [down] = ½(9.8m/s²)[itex]\nabla[/itex]t²
    and I know I have to isolate the "t" but how do I go about doing that, I keep getting it wrong

    The answer is 1.9s,
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jun 17, 2011 #2

    SteamKing

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    Re: Isolating "t" in horizontal motion equation

    Show us what you have done so far.
     
  4. Jun 18, 2011 #3
    Re: Isolating "t" in horizontal motion equation

    After posting this, I sat for another 2 hrs trying to figure out the problem and I finally understood it.

    18m [down] = ½(9.8m/s²)∇t²
    18(2)=9.8(t²)
    36=9.8(t²)
    36/9.8 = t²
    3.67=t²
    squareroot 3.67 = t
    1.91=t
     
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