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Throw a Pebble from a Pyramid

  1. Sep 12, 2009 #1
    1. The problem statement, all variables and given/known data

    A pebble is thrown off the side of a pyramide perpindicular to its slope. Find the vertical distance h under the assumption the drag forces are negligible.

    2. Relevant equations



    3. The attempt at a solution

    vx=vcos(90-Q) Q = theta
    vy=vsin(90-Q)

    ----------------

    x = vxt; therefore t = [tex]\frac{x}{v_x}[/tex] = [tex]\frac{x}{v_{x}cos(90-Q)}[/tex]

    -------------

    tanQ=[tex]\frac{y}{x}[/tex]; therefore y = xtanQ

    -------------

    y=yo + vyt - [tex]\frac{1}{2}[/tex]gt2

    y= [tex]\frac{xv_{x}sin(90-Q)}{v_{x}cos(90-Q)}[/tex] - [tex]\frac{9.8}{2}[/tex]([tex]\frac{x}{v_{x}cos(90-Q)}[/tex])2

    y= xtan(90-Q) - [tex]\frac{4.8x^{2}}{v^{2}cos^{2}(90-Q)}[/tex]


    -----------------
    Set both y's equal to each other

    xtanQ = xtan(90-Q) - [tex]\frac{4.8x^{2}}{v^{2}cos^{2}(90-Q)}[/tex]

    tanQ = tan(90-Q) - [tex]\frac{4.8x}{v^{2}cos^{2}(90-Q)}[/tex]

    now im gonna solve it for x

    x = [tex]\frac{v^{2}cos^{2}(90-Q)[tanQ-tan(90-Q)]}{-4.8}[/tex]

    and now i pluf it into y = xtanQ

    y = [tex]\frac{v^{2}tanQcos^{2}(90-Q)[tanQ-tan(90-Q)]}{-4.8}[/tex]

    seems a little to messy, did i miss somethig
     

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  3. Sep 13, 2009 #2

    rl.bhat

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    It is right. But you can put it in a better form.
    write tan(90 - Q) = cot Q
    cos(90 - Q) = sin Q
     
  4. Sep 13, 2009 #3

    ehild

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    Gold Member

    Take care, the slope goes downwards so y = -xtanQ. And 9.8/2 =4.9.

    ehild
     
  5. Sep 13, 2009 #4
    -tanQ = tan(90-Q)-[tex]\frac{4.9x}{v^{2}cos^{2}(90-Q)}[/tex]

    x = [tex]\frac{v^{2}cos^{2}(90-Q)}{4.9}[/tex][tan(90-Q) + tanQ]

    x = [tex]\frac{v^{2}sin^{2}Q}{4.9}[/tex][cotQ + tanQ]

    y = -[tex]\frac{v^{2}tanQsin^{2}Q}{4.9}[/tex][cotQ + tanQ]


    so there is nothin left to do with this one
     
  6. Sep 13, 2009 #5

    rl.bhat

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    In the last step put cotQ = 1/tanQ and simplify.
     
  7. Sep 13, 2009 #6
    y = [tex]\frac{-v^{2}sin^{2}Q[1 + tan^{2}Q]}{4.9}[/tex]
     
  8. Sep 13, 2009 #7

    rl.bhat

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    1 + tan^2θ = sec2θ
     
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