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Kinematics-motion in a straight line

  1. Oct 25, 2009 #1
    1. The problem statement, all variables and given/known data

    A dog sees a flowerpot sail up and then back past a window 5.0ft (1.77m) high. If the total time the pot is in sight is 1s, find the height above the top of the window to which the pot rises.


    2. Relevant equations

    standard set of kinematics equations are given;

    V=Vo+at
    V^2=Vo^2+2a(X1-X0)
    X=Xo+Vot+(1/2)at^2
    X=(1/2)(Vo+V)t
    X=Vt-(1/2)at^2


    3. The attempt at a solution

    Using the last equation given above find the velocity at the top of the window

    1.77=V(0.5)-(0.5)(-9.8)(0.5^2) solves for
    V=1.09m/s

    using equation 2 to calculate the distance traveled by the pot after passing the window

    0^2=(1.09^2)+(2)(-9.8)(X) solves for
    X=0.0606m

    The answer is 1.91cm. I've tried other ways too, but I can't seem to come up with the solution. Please help!
     
    Last edited: Oct 26, 2009
  2. jcsd
  3. Oct 25, 2009 #2
    Carefully look at your units. You're mixing feet and meters when you eventually plug your numbers in.
     
  4. Oct 26, 2009 #3
    I went through carefully but I can't find any error with my units, I don't see how this would be possible anyways as I never use any data given in feet, only meters.
     
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