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Kinematics problem two stones falling one with velocity one at rest

  1. Nov 16, 2005 #1
    hi please can sum one help me,

    a stone is dropped from he top of a cliiff at rest another stone it thrown downwards from the same point at 11ms-1. The two stones land at the same time, what is the height of the cliff,:frown:

    how do u work this out i have no idea,:cry:

    thankyou in advance

    :smile: bark00:smile:
     
  2. jcsd
  3. Nov 16, 2005 #2

    Päällikkö

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    Which equations do you think you're supposed to use?
     
  4. Nov 16, 2005 #3
    thankyou for helping me,

    im not sure,
    ive tried using combinations of v=u+at and s=1/2 (u+v)t but i dont know where to begin.
     
  5. Nov 16, 2005 #4
    This is a gravity problem...Can you set up the equations for both velocity and position in the horizontal (x) and vertical (y) direction ?


    ok, before actually giving specific information, i suggst you read this tutorial

    The second formula you gave is not known to me. Are you sure of it ?
     
  6. Nov 16, 2005 #5

    Galileo

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    Are you sure this is the question?
    So you release one stone from rest and you throw the other one vertically down at 11 m/s from the same cliff and they land omn the ground at the same time? How's that possible?

    Either I misunderstood the question, you copied it wrong or the height is zero and there would be no cliff.
     
  7. Nov 16, 2005 #6
    hi thankyou all for the help i think i go it now i got the awnser as 128.6 metres approx
     
  8. Nov 16, 2005 #7
    sorry missed out second one was trown a second later.
     
  9. Nov 17, 2005 #8

    dx

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    The second formula is equivalent to [tex] s = V_{avg}t [/tex] for constant acceleration.
     
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