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Homework Help: A boulder Rolling - Min Speed

  1. Feb 3, 2009 #1
    1. The problem statement, all variables and given/known data

    a 76 kg boulder is rolling horizontally at a top of a vertical cliff that is 20m above the surface of a lake. the top of the vertical face of a dam is located 100 m from the foot of the cliff, with the top of the dam level with the surface of the water in the lake. a level plain is 25 m below the top of the dam. what must the minimum speed of the rock be just as it leaves the cliff so that it will travel to the plain without striking the dam?

    2. Relevant equations

    3. The attempt at a solution

    20= 4.9 * t^2
    20/4.9= t^2
    t=square 20/49

    100 = vx0 * t
    100/2.02 = vx0
    vx0 = 45.45 m/s

    20+25 = .5 * 9.8 * t^2
    45 = 4.9* t^2

    V0x * 3.3 = 150m
    45.45*3.3 = 150
    150-100 = 50m

    I am not getting the min speed. Is it the 45.45? I believe so, but the program doesnt. Appreciate any help. The 50 m is correct.
  2. jcsd
  3. Feb 3, 2009 #2

    Doc Al

    User Avatar

    Staff: Mentor

    Redo that last bit of arithmetic.
  4. Feb 3, 2009 #3
    49.5, thanks. Have had very little sleep.
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