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The Drowsy Cat - Kinematics

  1. Dec 11, 2004 #1
    The Drowsy Cat - Kinematics!!!!!!!!

    Can anybody help me get started? I have been working on this problem forever....
    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 0.50s and the top to bottom height of the window is 2.00m. How high above the window does the flowerpot go? The answer is 2.34m but I don't know how to get it, I know that it is a kinematics free fall question but I keep getting too many variables. If anyone knows how to go about answering this, that would be awesome! Thank you!
  2. jcsd
  3. Dec 11, 2004 #2


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    The pot's arc will be symmetrical, that means it will take exactly half the time (0.25 seconds) to travel upwards the 2m. (The second half of the pot's trajectory is irrelevant.) Does that simplify things?
  4. Dec 11, 2004 #3
    Yes! Thank you very much. So in these problems, then, can you always assume that the time to go up = the time to come back down? (assuming an ideal free fall situation)
  5. Dec 11, 2004 #4


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    For an object thrown from the ground, its velocity on the way down will be equal in magnitude but opposite in direction then at the same height on the way up. So yes it will take the same amount of time to pass the window going going up as going down.

    If you set up the falling body equation of motion with t=0 and x=0 at the bottom of the window it works out nicly.
  6. Dec 12, 2004 #5


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    My stars & garters - I helped someone! :-)
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