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2-dimential motion

  1. Sep 19, 2009 #1
    1. The problem statement, all variables and given/known data
    A placekicker is about to kick a field goal. The ball is 26.5 m from the goalpost. The ball is kicked with an initial velocity of 21.5 m/s at an angle θ above the grojund. Between what two angles, θ1 and θ2, will the ball clear the 2.81-m-high crossbar? Give your answers as (a) the smaller angle and (b) the larger angle.(Hint: The following trigonometric identities may be useful: sec θ=1/(cosθ) and sec2θ=1+tan2θ


    2. Relevant equations
    x=26.5m
    y=2.81m
    v0=21.5m/s
    ax=0 m/s
    ay=-9.8 m/s

    d=v0t+1/2at2


    3. The attempt at a solution
    Okay I'm stuck. I believe I should separate the initial velocity into an x and y component. Possibly 26.5cosθ=the x component and 26.5sinθ=the y component.
    Then maybe plug in those values into the equation to find time? Can someone tell me if I'm anywhere near the right track? The farther along I go the bigger mess of trig functions I get
    Please
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Sep 19, 2009 #2

    Redbelly98

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    Welcome to Physics Forums :smile:

    Yes, exactly.

    I want to say yes, but it depends which values and which equation, specifically, you are talking about.
     
  4. Sep 19, 2009 #3
    So...
    26.5=(21.5cosθ)t
    and
    2.81=(21.5sinθ)t+1/2(-9.8)t^2

    is that the right next step? Solve the top equation for t and plug that t into the second equation to be able to solve for θ?
     
  5. Sep 19, 2009 #4

    Redbelly98

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  6. Sep 19, 2009 #5
    Okay that sound good in theory :). But my trig isn't that good so i get a little confused.
    So far I have
    26.5/(21.5cosθ)=t
    Plugged it in to get:
    2.81=(21.5sinθ)(26.5/21.5cosθ)+(-4.9)(26.5/21.5cosθ)^2
    Can I get some hint at what I should do next to be able to get rid of all the trig functions
     
  7. Sep 19, 2009 #6

    Redbelly98

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    Sure!

    One term contains sinθ/cosθ, which is equal to ___θ?

    The other term has (1/cosθ)^2 ... use the two trig identity hints from the problem statement to change that.
     
  8. Sep 19, 2009 #7
    Okay. Using that this is what I'm getting:
    2.81=26.5tanθ+(-6.04*(1+tan2θ))
    So I should distribute?
    2.81=26.5tanθ-6.04-6.04tan2θ
    Put everything to one side?
    0=-6.04tan2θ+26.5tanθ-8.85
    What should I do now?
     
  9. Sep 19, 2009 #8

    Redbelly98

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    Well, if we let x = tanθ, the equation looks like this:

    -6.04x^2 + 26.5x - 8.85 = 0​

    Can you solve that for x?
     
  10. Sep 19, 2009 #9
    Oh the quadratic.
    Ok so....
    -26.5+√702.25-(4*-6.04*-8.85)
    -------
    2*-6.04
    and the same but with a minus
    -26.5-√702.25-(4*-6.04*-8.85)
    -------
    2*-6.04
    so x=0.36 and 4.02
    Then I plug it back in right? So I get:
    tan-10.36=θ and tan-14.02=θ
    So θ=19.8 and 76.03?
     
  11. Sep 19, 2009 #10

    Redbelly98

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    That should work out, but I'm seeing something wrong when I check these angles against these equations:

    Okay, so using the 19.8 degrees solution, I solve the 1st equation for t and get 1.31 s.
    Then using 19.8 degress and 1.31s in the 2nd equation, I get 1.13 instead of 2.81. So there is a mistake somewhere ...

    I get something different than the 6.04 number ... I would double-check how you got that.

    You have the right method to work this through, it's just a matter of correcting the arithmetic/algebra error.

    I'm logging off the computer for tonight. When you work this through and get the angles, remember to check each angle in the equations:
    Good luck!
     
  12. Sep 19, 2009 #11
    Okay I got it this time for sure :). Your right my 6.04 answer should have been 7.44 just a miscalculation. Once I worked it through I got 23.8 and 72.2 and those work in the equations :)
    Thank you for your help
     
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