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Homework Help: Falling cat - how long is the cat in the air

  1. Nov 19, 2007 #1
    The problem says A cat falls from a tree, dropping 16 meters to the ground. How long is the cat in the air?

    Im not confused with the problem, Im confused with the math..The book gives me of course the gravitational constant of earth and all that. The thing is, earlier it gave me an equation that is used here, which is s=initial velocity(time)+1/2(acceleration)(time) squared (Sorry, Im not sure how to use equations on this thing..). The book says that you would get the answer by using t=square root of 2s/(gravity).

    So...I dont understand exactly why it goes from distance=1/2(gravity)(time)squared to the equation time=square root of 2(s)/(gravity). I understand that you are looking for the time, not the distance, so those two switch out, but what happens to the 1/2 part of this equation? Is it there automatically due to the 2(s)/(gravity) part?

    Im sorry if I didnt explain this well enough for anyone to understand what I mean, and if thats the case please tell me and Ill try to clarify...
  2. jcsd
  3. Nov 19, 2007 #2


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    So basically the question is how to rewrite the equation
    [tex]s = \frac12 g t^2[/tex]
    [tex]t = \sqrt{ \frac{ 2s }{ g }[/tex]

    What we can use for this is the balance method, which basically says that you can do any operation you wish as long as you apply it equally on both sides (there are exceptions but in general this works for any basic operation).
    We want to isolate the t on one side, so first let us get rid of the other terms: multiply both sides by 2 (divide by 1/2):
    [tex]2 s = 2 \frac12 g t^2 = g t^2[/tex]
    and divide by g
    [tex]\frac{2 s}{g} = \frac{g}{g} t^2 = t^2 [/tex]
    and now take the square root on both sides:
    [tex]\sqrt{\frac{2s}{g}} = \sqrt{t^2} = t[/tex]
    (technically, only if t is positive but in this problem it is).

    Do you follow or is this Japanese (or if you happen to be Japanese, some other language like Greek) to you?

    PS, I just found this page -- if you feel you're having trouble solving this sort of equations try a few, maybe it will help you pinpoint the problem.
    Last edited: Nov 19, 2007
  4. Nov 19, 2007 #3
    Hmm..ahh I get it. Algebra 1 is coming back to me haha. Thank you very much that helped alot. My book here doesnt explain that so I got confused as to why we'd change the equation to a square root from the original, but now I see thats how it would be answered by following this method. Thanks.
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