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I must find Time.

  1. Sep 2, 2007 #1
    1. The problem statement, all variables and given/known data 2 Identical pellet guns are fired simultaneously from the edge of a cliff. These guns impart an initial speed of 30.0 m/s to each pellet. Gun A is fired straight upward, with the pellet going up the cliff and then falling back down, eventually hitting the ground beneath the cliff. Gun B is fired straight downward. In the absence of air resistance, how long after pellet B hits the ground does pellet A hit the ground.
    a= 9.8 m/s^2
    Vo=30.0 m/s

    2. Relevant equations

    y=Vot + 1/2 at^2
    or t=Vo*2/a
    v=Vo + at

    3. The attempt at a solution

    Here is what I did. 30.0 m/s *2/9.8 m/s^2= 6.12 s. The answer is right, but did I do the problem right.
  2. jcsd
  3. Sep 2, 2007 #2


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    How did you come up with that?
  4. Sep 2, 2007 #3


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    Well, [tex]ms^{-1} [/tex] times [tex] ms^{-2} [/tex] would not give an answer in seconds. Since both would fall from the cliff with an initial velocity of [tex] 30 ms^{-1}[/tex] and with the same acceleration, the time difference would be the time taken for pellet A to go up and come back down to the level of the cliff.
  5. Sep 2, 2007 #4


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    No it's the right answer... the units come out right. it's ms^-1 divided by ms^-2.
  6. Sep 2, 2007 #5
    so did i do it right
  7. Sep 2, 2007 #6


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    You have not told us what you did to arrive at the expression (in particular the lefthand side of the expression):

    "30.0 m/s *2/9.8 m/s^2= 6.12 s"

    Thus we can not yet tell you if what you did was right.
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