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Free-fall Acceleration Problem

  1. Oct 5, 2008 #1
    1. The problem statement, all variables and given/known data

    A drowsy cat spots a flowerpot that sails first up and then down past an open window. The pot is in view for a total of 0.50 s, the top-to-bottom height of the window is 2.00 m. How high above the window top does the flowerpot go?


    2. Relevant equations

    Constant acceleration equations


    3. The attempt at a solution

    The answer in the back of the textbook is listed as 2.34 m above the window top. However, using the following technique:

    v (velocity at the top)


    v= v0 + at
    0= v0 + at
    0 - at = v0
    0 - (-9.8 m/s^2)(.5) = v0
    4.9 m/s = v0

    v^2 = v0^2 + 2a(x-x0)
    0 = v0^2 + 2g(y-y0)
    -v0^2 = 2gy
    -v0^2/2g = y
    = 1.225 m

    which apparently is inconsistent with the text's answer. Help!!!
     
  2. jcsd
  3. Oct 5, 2008 #2

    Doc Al

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    Staff: Mentor

    Not sure what you are doing here. What is v0 supposed to be? Why is the time set at .5 seconds? (You found the initial speed of something that rose for .5 seconds before falling back down.)
     
  4. Oct 5, 2008 #3
    v0 is the initial velocity of the flowerpot before it rose to its highest point, v, and then descended.

    the problem indicated that .5 is the total time it was in view, which i'm assuming is including the time the "cat" observed the flowerpot both rise to the top point of the window and then subsequently the time it observed the flowerpot descent from the top point of the window to the bottom of the window.

    the problem is oddly worded, so i may be getting it very wrong...
     
  5. Oct 5, 2008 #4

    Doc Al

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    Staff: Mentor

    I presume you mean the speed it has as it passes the bottom of the window?

    The flowerpot rises to some unknown height above the top of the window before coming back down again, so the time it takes it to go from the bottom of the window to its highest point is not necessarily .5 seconds. (But must be greater than .25 seconds.)
     
  6. Oct 5, 2008 #5
    so how would you say i'd approach this problem?

    i'm trying to find a point of departure, but i still do not understand how to work with the data given. so i take it .5 seconds is the time it takes both to ascend and descend?
     
  7. Oct 5, 2008 #6
    I'm going to jump in and try to help you along. You won't be able to do this in one easy step, so let's look at what you have.

    At the bottom of the window the flowerpot has some velocity v0. It goes up, above the window, at some top point has a final velocity of 0, then comes back down and passes by the window again. Since the total time the cat can see the flowerpot is .5 seconds, he sees it for .25 seconds going up, then .25 seconds coming back down. We also know that the height of the window is 2.0m. And you already know that the acceleration is 9/8 (or -9.8)m/s/s. To start out, what can you find out if you know time, acceleration and distance?
     
  8. Oct 5, 2008 #7
    velocity?
     
  9. Oct 5, 2008 #8
    Yes, just be sure you understand at what point in the ascent (or descent) the velocity represents.

    Once you have that, you should be able to use that information to find the answer you are looking for.
     
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