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Initial Velocity (Without time and acceleration)

  1. Sep 16, 2009 #1
    1. The problem statement, all variables and given/known data
    You are standing at the top of a cliff that has a stair step configuration. There is a vertical drop of 7 m at your feet, then a horizontal shelf of 9 m, then another drop of 3 m to the bottom of the canyon, which has a horizontal floor. You kick a 0.26 kg rock, giving it an initial horizontal velocity that barely clears the shelf below.

    What initial horizontal velocity v will be required to barely clear the edge of the shelf
    below you? How far from the bottom of the second cliff with the rock land?

    2. Relevant equations

    d=Vi(t)+(1/2)(a)(t^2)

    3. The attempt at a solution
    Cant seem to get around the problem without knowing more.
     
    Last edited: Sep 16, 2009
  2. jcsd
  3. Sep 16, 2009 #2

    kuruman

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    Can you find the time required for the rock to reach the point where it barely clears the edge of the shelf?
     
    Last edited: Sep 16, 2009
  4. Sep 16, 2009 #3
    I havent been able to
     
  5. Sep 16, 2009 #4
    You can find the time by using gravity and the fact that the first step is 7 meters down. How long does it take for the rock to fall 7 m? If you find it you can calculate how fast it must travel in a horizontal direction.
     
  6. Sep 16, 2009 #5
    Could you help me out a little more?
     
  7. Sep 16, 2009 #6
    So im assuming you mean:
    7m/9.8=.7142857s to make it to the bottom =t

    using

    d=(Vi+Vf/2)(t)
    therefore
    9m=(0+Vf/2)(.7142857s)
    Vf=25.2

    that not right?
     
  8. Sep 16, 2009 #7
  9. Sep 16, 2009 #8

    kuruman

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    Use the equation that you initially quoted as "relevant" for the vertical motion. Identify ll the quantities - you know all of them except the time. Remember, vertical motion only.
     
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