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Kinimatics check

  1. Apr 13, 2013 #1
    I have gotten the right ans for this question, but I am not too sure if I am worked it out the right way.

    Question: A particle is projected vertically upwards from a point O with speed [itex]u ms^{-1}[/itex]. Two seconds later it is still moving upwards with a speed [itex]\frac{1}{3}u ms^{-1}[/itex]. Find a: the value of u, b: the time from instant that the particle leaves O to the instant that it returns to O.

    I worked out U which is 29.4ms^-1.

    It b: I am having trouble with, as I stated I have the right ans, but to sure on if I have gotten the ans correctly.

    What I have done it is take the part where the ball is at 1/3 u and workout out the time of travel from the part where it passes 1/3u again and the add on 4 seconds. Not very good at explain hopefully the math will show better.

    [itex]0=9.8t-4.9t^2 → 4.9t^2-9.8t=0 [/itex] solve and t=0 and t=2. So I then add on the 4sec and total time is 6 secs. Is this correct or not?

    I would appreciate any help, big thanks in advance.

    Sorry for the repost, the first one keeps telling me its invalid I cant edit it at all. So could one of the mods remove my older post.
     
  2. jcsd
  3. Apr 13, 2013 #2

    Doc Al

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    Staff: Mentor

    Please explain where this equation comes from. Where did you get a velocity of 9.8?

    Why not deal directly with velocity equations? Much simpler.
     
  4. Apr 13, 2013 #3
    I got it from the 1/3*29.4. I didn't know that I could deal with directly as there are two different velocity, for the way up. That is why I broke it down.
     
  5. Apr 13, 2013 #4

    Doc Al

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    Staff: Mentor

    But that kind of assumes the answer, doesn't it?

    Just use vf = vi - gt .
     
  6. Apr 13, 2013 #5
    Okay I am with you. Thanks for the help.
     
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