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Finding initial speed given two angles and a distance

  1. Sep 22, 2009 #1
    1. The problem statement, all variables and given/known data

    A child runs down a 11 degree hill and then suddenly jumps upward at a 16degree angle above the horizontal and lands 1.2 m down the hill as measured along the hill

    What was the child's initial speed?

    2. Relevant equations

    Projectile motion

    x = Vx*t

    y = yo + Vyo*t - 1/2gt^2

    3. The attempt at a solution

    tried solving for time t from the vertical motion equation y = yo+ Vyo*t - 1/2gt^2 and plug it in the
    x = vx*t..but it doesnt work
     
    Last edited: Sep 22, 2009
  2. jcsd
  3. Sep 22, 2009 #2

    rl.bhat

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    Hi heyhowsitgoin
    Welcome to PF.
    If you draw the figure of inclined plane and the landing of the child, you can see that
    y = 1.2*sin11.
    t = x/vx = 1.2*cos11/vo*cos16
    vyo = vo*sin16.
    Substitute these values in the equation and solve for vo.
     
  4. Sep 23, 2009 #3
    Hi thanks for the reply. Which equation are you referring to solve for vo?
     
  5. Sep 23, 2009 #4

    rl.bhat

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    y = yo+ Vyo*t - 1/2gt^2
     
  6. Sep 23, 2009 #5
    hrmm ok

    so i end up with 0.229 = 0 + Vo*sin16 (1.2*cos11 / Vo*cos16) - 1/2(9.8)(1.2*cos11/Vo*cos16)^2

    is this right? not sure how to solve for Vo at this point.
     
  7. Sep 23, 2009 #6

    rl.bhat

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    It is right. Now simplify everything. And solve for Vo.
     
  8. Sep 23, 2009 #7
    Hello,

    I was looking at this question and was not able to isolate Vo in the final equation. I ended up with a negative number under a square root...
    Any suggestions on how to correctly isolate Vo?
     
  9. Sep 23, 2009 #8

    rl.bhat

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    In the equation final position y is zero, and initial position yo is 0.229
     
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