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1-D Kinematics Equations involving time & distance

  1. Aug 21, 2011 #1
    Hey, I've been working on this problem for quite a while now, and I can't seem to come up with the solution. I would love some help in the right direction.

    1. The problem statement, all variables and given/known data
    Someone spots a flowerpot that sails first up and then down past an open window. The pot was in view for a total of 0.61 s, and the top-to-bottom height of the window is 2.25 m. How high above the window top did the flowerpot go?

    ttotal= .61 s
    Height = 2.25 m

    2. Relevant equations
    Vf = Vi + axt
    Xf = Xi + 1/2(Vi + Vf)t


    3. The attempt at a solution
    To try and solve it, I figured that if the pot was in view for .61s total, it was .305s on each trip. So I figured its Vi = 2.25m/.305s = 7.377 m/s. So if its Vi=7.37, then it would take .7527 seconds to slow to 0m/s due to -9.8m/s2 for gravity. After .7527 seconds it would fly to 2.77m, making it .53 meters above the window. Unfortunately, this is not the correct answer. Any suggestions?
     
  2. jcsd
  3. Aug 21, 2011 #2

    lewando

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    Gold Member

    You calculated an average velocity that is neither vbot or vtop. Since you know a lot about x and t, why not use x = x0 + vx0t + axt2/2, (coordinate system: +x going up).
     
  4. Aug 21, 2011 #3
    Plugging my knowns into that equation still comes out to be the same thing though.

    X = 0 + (7.377 * .7527) + (-9.8 * (.7527 ^ 2) * .5)
    X = 0 + (5.55) + (-2.776)
    X = 2.776
    2.776 - 2.25 = .53 meters, which is still wrong.
     
  5. Aug 21, 2011 #4
    Is my problem coming from the fact that I can't actually split the .61s in half for each trip because they wouldn't be evenly distributed, making my initial velocity wrong? How do I fix that?
     
  6. Aug 21, 2011 #5

    gneill

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    Staff: Mentor

    No, that's not your problem; The trajectory is symmetrical so it's split half and half as you've done.

    Perhaps concentrate on finding the required velocities Vbot and Vtop to begin with. You know that on the way up the pot will have some "initial" velocity at the bottom (Vbot) and must traverse the span of the window (known) in a given time (known). What equation governs the distance with respect to time?
     
  7. Aug 21, 2011 #6
    So, trying that:

    2.25 = 0 + Vi * .305 + (.5 * -9.8 * (.305 ^ 2))
    2.25 = .305Vi + (-.455)
    2.70 = .305Vi
    Vi = 8.87 m/s

    t = -8.87 / -9.8 = .905 sec

    Xf = 0 + 1/2 * (8.8 - 0) * .905 = 3.982 m

    3.982m - 2.25m = 1.732 m

    Which is still wrong. Ugh.
     
  8. Aug 21, 2011 #7

    gneill

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    Staff: Mentor

    Okay so far.
    Whoa! What equation is that? Why not use [itex]d = v_o t - \frac{1}{2}g t^2[/itex]?

    Alternatively, use energy conservation to find the height :wink:
     
    Last edited: Aug 21, 2011
  9. Aug 21, 2011 #8
    So plugging into that one:

    d = 8.8 * .905 - (.5 * -9.8 * (.905^2)) = 11.97 m
    11.97 m - 2.25 m = 9.7m.

    Which is still wrong. Thanks for the help Gneill, sorry Im so difficult.
     
  10. Aug 21, 2011 #9

    gneill

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    Staff: Mentor

    The "-" sign in the equation was intended to make the acceleration negative. g is a positive constant (9.807 m/s2). The double negative has made the calculation invalid.
     
  11. Aug 21, 2011 #10
    So instead, its:

    7.964 - .5 * 9.8 * (.905^2) = 3.950

    3.950 - 2.25 = 1.7 meters above the window?
     
  12. Aug 21, 2011 #11

    gneill

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    Staff: Mentor

    You seem surprised. Why don't you verify by checking the velocity of a pot dropped from that height when it reaches the bottom sill of the window? :smile:
     
  13. Aug 21, 2011 #12
    1.7 still came back as incorrect :P
     
  14. Aug 21, 2011 #13

    gneill

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    Staff: Mentor

    You may need to retain a few more decimal places in your intermediate results and supply another significant figure in your answer.
     
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