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Calculate when displacement are equal

  1. Feb 28, 2013 #1
    1. The problem statement, all variables and given/known data

    One stone is dropped from the the top of a tall cliff, and a second stone with the same mass is thrown vertically from the same cliff with a velocity of 10.0 m/s [down], 0.50seconds after the first. Calculate the distance below the top of the cliff at which the second stone overtakes the first?

    2. Relevant equations
    v=0
    d= -1/2(a)(t)^2

    v=10
    d=10(t)-1/2(a)(t)^2

    3. The attempt at a solution

    we have two unknowns, how can we solve it?!?!
     
  2. jcsd
  3. Feb 28, 2013 #2

    SammyS

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    The time in this equation should be (t - 0.50), if t is the time in the previous equation .
    Two equations in two unknowns.
     
  4. Feb 28, 2013 #3
    i still don't get the answer, can you explain the steps please
     
  5. Feb 28, 2013 #4

    SammyS

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    First rewrite the second equation,

    d=10(t)-1/2(a)(t)2 ,

    using (t - 0.5) in place of t. (This is because the second stone is in the air for 1/2 second less time than the first stone.)

    Then "FOIL" the (t - 0.5)2, multiply out everything in the expression and then collect terms.

    See what you have then.
     
  6. Feb 28, 2013 #5
    so after you foil, you make the displacement equal to each other, and then solve for t, and then plug "T" into the original equation?
     
  7. Feb 28, 2013 #6

    SammyS

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    Yes.
     
  8. Feb 28, 2013 #7
    can you also do: d= -1/2(a)(t+.5)^2

    cas the ffirst stone is .5 seconds in air longer than the second stone
     
  9. Feb 28, 2013 #8

    SammyS

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    You can do either

    t for the first stone and (t - 0.5) for the second stone,

    OR

    (t - 0.5) for the first stone and t for the second stone,

    but not both (t + 0.5) and (t - 0.5).
     
  10. Feb 28, 2013 #9
    don't you mean "t for the first stone and (t "+" 0.5) for the second stone,"?
     
  11. Feb 28, 2013 #10

    SammyS

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    No. No matter how you express the time for each stone the time for the second stone is always 0.5 s less than the time for the first stone.

    Therefore, the time for the first stone is always 0.5 s more than the time for the second stone.
     
  12. Feb 28, 2013 #11
    (t - 0.5) for the first stone and t for the second stone,

    for that cas the time for the first one is 5 less than the time for the second stone.

    so is this correct "(t + 0.5) for the first stone and t for the second stone,"?
     
  13. Feb 28, 2013 #12

    SammyS

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    Then, while the first stone falls for say 1 second, the the second stone would fall for 1 + 0.5 seconds. That's just not correct.
     
  14. Feb 28, 2013 #13
    ohh lol, now i get it, thanks :D
     
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