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Minimum clearence wiith range

  1. Feb 20, 2005 #1
    You are standing at the top of a cliff that has a stairstep configuration. There is a vertical drop of 6 m at your feet, then a horizontal shelf of 6 m, then another drop of 4m to the bottom of the canyon, which has a horizontal floor. You kick, a .72 kg rock,giving it an inital horizontal velocity that barely clears the shelf below. The acceleration of gravity is 9.8m/s^2. Consider air friction to be negligible.

    What initial horizontal velocity veloctiy v will be required to barely clear teh edge of the shelf below you? Answer in units of m/s.

    sorry this one i dont really haev a clue on what should i be looking at?
     
  2. jcsd
  3. Feb 20, 2005 #2

    jamesrc

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    You can solve this by calculating how long it takes a projectile (with 0 initial vertical velocity) to drop 6m (the drop to the ledge). Then use that time to find out what initial horizontal velocity it would take to move 6 m (the shelf length) in that time. Remember that since there is no air resistance, the horizontal velocity is a constant (not accelerated). You could then find out where the rock ultimately lands, if asked.
     
  4. Feb 20, 2005 #3
    are you saying then that because its the same size 6 x 6 that the velocity in each direction is teh same? but also to calculate the velocity in y isnt it just 6/9.8?
     
  5. Feb 22, 2005 #4

    jamesrc

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    No to both questions.

    The velocity in the horizontal direction is what you need to find. The "initial" horizontal velocity is the same as the final horizontal velocity since it is not accelerated in that direction.

    The velocity in the y direction is not needed in this problem per se, but if you want to find it, it is a function of time: vy = at, where a = -9.81m/s/s (I've defined "up" as positive for this problem.) and t is the time elapsed since the kick. If, say, you want to find the y-velocity when the rock reaches the ledge, you could use vy2 = voy2 + 2aΔy, where voy is the initial y-velocity (0 m/s), a =-9.81m/s/s again, and Δy is the final height minus the initial height = -6m.

    You could use that result to find the time elapsed using v = at. Or you could have found the time directly using y -yo = .5*a*t2 + voyt

    Either way, the initial horizontal velocity is found by dividing the distance moved in the horizontal direction (+6m) by the time elapsed, t.
     
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