1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Help with Problem (Complicated Kinematics)

  1. Aug 24, 2006 #1
    Help with Problem plz (Complicated Kinematics)

    A drowsy cat spots a flowerpot that sails first up and then down past an open window. The pot was in view for a total of 0.35 s, and the top-to-bottom height of the window is 1.95 m. How high above the window top did the flowerpot go?

    -At first I tried finding the velocity by m/s. Then later when I thought about it, the velocity is not constant in this problem. Now I have no clue what to do, even though I know 9.8 m/s^2 as gravity and the velocity in the distances of the window.
  2. jcsd
  3. Aug 24, 2006 #2

    Doc Al

    User Avatar

    Staff: Mentor

    You'll need to attack this problem in steps. There are several ways to go. Try figuring out the speed of the flowerpot at various points. Hint: Start by finding the average speed of the flowerpot as it passes by the window.
  4. Aug 24, 2006 #3
    Think about the problem. Will the velocities up and down be the same? If so, then you can calc the initial and final velocities of the pot going past the window, and the rest is easy peasy. Try drawing a diagram also.
  5. Aug 25, 2006 #4


    User Avatar
    Homework Helper

    The up and down speeds of the pot is the same at the same height above the ground, they are just in opposite directions. This can be seen from the equation

    [tex]v^2 = u^2 - 2gy[/tex]

    which do not distinguish between the up and down motions. This means that the up and down times to cross the window will be the same when one considers

    [tex]v = u - gt[/tex]

    for the separate cases of the up and down motion past the window. For instance say the velocity at the bottom is va and the top vb, then for the going up case (taking up as positive) we have

    [tex]v_b = v_a - gt[/tex]

    and for the going down case (taking down as positive) we have

    [tex]v_a = v_b + gt[/tex]

    From this info you can calculate va from

    [tex]y = v_at - 4.9t^2[/tex]

    But I must say I see several logical errors in the problem:
    1. Why would someone want to throw a flower pot upwards? He would just have to catch it again on its way down or jump out of the way.
    2. How or why would the cat measure the time for the pot to cross the window?
    3. Who taught him to add?
    4. How did he communicate his results to the physicist?
    5. If the cat did not do it, why is he mentioned in the problem anyway?
    Last edited: Aug 26, 2006
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Help with Problem (Complicated Kinematics)