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How to find angle whit system of equations

  1. Aug 21, 2012 #1
    1. The problem statement, all variables and given/known data
    i kick a ball with an initial velocity of 18m/s. there is a wall at 8m that has a height of 1.5 meters. whith what angle should i kick the ball so that it bearly passes the wall?


    2. Relevant equations
    first equation: hf= 1/2gt^2+vit+hi
    second equation: V=Δx/t

    3. The attempt at a solution
    so what i am trying to do is a system of equations isolating t in the second equation and placing it in the first equation. in velosity in the y direction i use 18sin(θ) and for v in the x i use 18cos(θ).
    Also as final heigth i am using 1.6m(just a little higher than the wall) and as xfinal i am using 8.1m(also bearly passing the wall)

    the problem is i cant soleve the equation. is this the right way to do it?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Aug 21, 2012 #2

    CAF123

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    Gold Member

    So you have [tex] s_y = v_{oy}t -\frac{1}{2}gt^2,[/tex] and you have values for [itex] s_y [/itex] and [itex]v_{oy}. [/itex]
    If you know the wall is 8m away, and you know [itex]v_{ox}, [/itex] then by rearranging [itex] s_x = v_{ox}t [/itex] for [itex] t [/itex] and subbing into the above eqn, you should be able to solve for [itex] θ.[/itex]
    It is just a case of rearranging the final eqn for [itex] θ. [/itex] Is this the part you are having difficulty with?
     
  4. Aug 21, 2012 #3
    sorry what is sy(changes in y?) , and yes that is where i am having trouble. i end up with cosines and sines and cant unite them or cancel them or anithyng
     
  5. Aug 22, 2012 #4

    ehild

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    It is the right way in principle but "just a little higher" can mean anything. Why not 1.55 m? or 1.51 m? So choose y=1.5 m at x=8 m.
    After eliminating t, you get an equation which contains both the tangent and the cosine of theta. (y=tan(θ)x-(g/2)x2/(182cos2(θ)). You certainly know that cos2(θ)=1/(1+tan2(θ)). Use that relation and you get a quadratic equation for tan(θ).

    ehild
     
    Last edited: Aug 22, 2012
  6. Aug 22, 2012 #5

    CAF123

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    [itex] s_y [/itex] is the vertical displacement of the object. I implicitly assumed [itex] s_o = 0 [/itex] in the equation [tex] s_y - s_o = v_{oy}t -\frac{1}{2}gt^2 [/tex]
     
  7. Aug 22, 2012 #6
    now i got it, great explenation! thanks
     
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