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One-Dimensional Kinematics Problem

  1. Sep 5, 2006 #1
    Hey guys,

    I just joined and I have a question. Here is my problem:

    Sitting in a second-story apartment, a physicist notices a ball moving straight upward just outside her window. The ball is visible 0.291 s as it moves a distance of 1.22 m from the bottom to the top of the window. What is the greatest height of the ball above the top of the window?

    How long does it take before the ball reappears?

    This is what I have done so far & I have worked on this problem for days and I am ready to cry, seriously.

    I figured out the average velocity as 4.19 m/s.

    Then I figured out the velocity at the bottom of the window as 7.05 m/s. After that I try to use the formula to find the position as a function of time with the formula x= x_o + v_0t - .5gt^2. At this point I don't know what to do. Do I put in 0 for x_0 or do I put in 1.22? If I put in 0 and use 0.291 as the time I end up with x = 1.64 m. THat answer isnt right according to LON CAPA. ANd after I do figure out the position how do i figure out the time. Do I just divide the position by the 7.05? I need some major help. I need to know how to set up the last formula. Maybe I need to add my answer to the 1.22 or subtract it. I really don't know. Let me know if anyone can help me. Thank you.

    You can also send me an email if you dont want to post on the board.

    Last edited: Sep 5, 2006
  2. jcsd
  3. Sep 5, 2006 #2
    We must determine the velocity at the bottom of the window, v0, and the velocity at the top of the window, v.

    For two unknowns we need two equations:

    (v + v0)/2 = average velocity

    v2 = v02 + 2ax

    These hints may provide a good starting help for this problem. The other calculations are more accessible.

    Not correct.
  4. Sep 5, 2006 #3
    is the velocity at the bottom of the window 5.62 m/s and the velocity at the top of the window 2.76 m/s? Was the 4.19 m/s correct as the average velocity? Do i use 1.22 as x_0 in the position equation or do I just use 0?

    if i use 1.22 i get x=2.44, if i use 0 i get x=1.22. Are either of those correct as the greatest height? And if they are how do I figure out the time after that?
  5. Sep 5, 2006 #4
    anyone else have any advice on this one? the above advice didnt really help me. I need some step by step instructions or something. i have been working on this problem for too long.
  6. Sep 5, 2006 #5
    Using the 2 equations PPonte gave you, find the velocity at the top of the window. Then use this in the equation v2 = v02 + 2ax with v set to 0, as it is stationary at its greatest height, and v0 the velocity at the top of the window, and solve for x.

    Then you need t for travelling a distance 2x, so use the equation d=ut+0.5at2, and solve for t.
  7. Sep 5, 2006 #6
    I'm not eaxtly sure if this would work but you could use this equation to try and solve for time portion of your problem; V^2=Vi^2+2ax. You know the initial and final velocity and you the displacement is 1.22m. I would sove for acceleration and then use this equation; (Vf-Vi)/a=t. I also think your initial position is zero and the ball moves a total of 1.22m. Hopefully this will help.
  8. Sep 5, 2006 #7
    So I figured out the first answer. It is 0.391 m. Now I need to figure out the time. When i put it in the equation i get two answers...0.286 and 0.278...my equation is

    0.391 = 2.77t - .5(9.81)t^2

    What is the correct answer for the second part of the problem originally posted in the beginning?
  9. Sep 5, 2006 #8
    i tried 0.286 and it was incorrect and i only have one more try left.
  10. Sep 5, 2006 #9
    is it 0.572? that is twice 0.286
  11. Sep 5, 2006 #10
    One more try left?
    What are you refering to?

    If this is some homework assignment and you really have trouble understanding this even after the generous hints it is not going to help you by just getting the answers here and giving the teacher the right numbers. That way you are fooling the teacher and yourself.
    Last edited: Sep 5, 2006
  12. Sep 5, 2006 #11
    I have gotten hints from the teacher also. we have emailed back and forth. Before I did get help from him I tried a lot of numbers I came up with on LON CAPA, therefore, I only have one more try left, so I want to make sure I am doing it the right way. I know I am close and I have the right formula. I have been working on this for a while.
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