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Projectile problem

  1. Aug 20, 2009 #1
    1. The problem statement, all variables and given/known data
    A projectile is fired from a height of
    [tex]
    y_0
    [/tex]
    ft above a level terrain,with a velocity of
    [tex]
    v_0
    [/tex]
    and at an angle
    [tex]
    \alpha
    [/tex]
    with the horizontal.
    Find
    a. The equation of the particle's path.
    I have solved this:
    [tex]
    y=y_0+(\tan\alpha)x-\frac{g(\sec^2\alpha)}{2(v_0)^2}x^2
    [/tex]
    b.It's horizontal range-take the x axis on ground level.
    I have also solved this as the horizontal range is found from setting y=0
    this leads to a quadratic in x and taking the positive root gives the range.
    c. It's maximum height.
    Again I have solved this by setting dy/dx =0
    Finally
    [tex]
    y_{max}=y_0+\frac{v_0^2\sin^2\alpha}{2g}
    [/tex]
    d. When will it reach the ground?
    Again I have solved this.
    [tex]
    t=\frac{range}{v_0\cos\alpha}
    [/tex]
    The denominator is the horizontal component of vo.
    IT IS THE NEXT PART THAT I CANNOT SOLVE.
    e. What is the value of alpha that will make the range a maximum?
    John


    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Aug 20, 2009 #2

    tiny-tim

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    Hi John! :smile:
    ok, what is the range that you found, and that you now have to maximise?
     
  4. Aug 20, 2009 #3

    kuruman

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    Gold Member

    In part (d) what is "range"? The given quantities are v0, α and y0. Unless you assume that the projectile returns to the same height y0, you will not be able to proceed. If this assumption is valid, take y0=0 and proceed as tiny-tim suggests. But you do need to find an expression for the range in terms of the given quantities.
     
  5. Aug 20, 2009 #4
    OK the range is obtained by setting y=0.
    This leads to the following quadratic in x.
    [tex]
    \frac{g\sec^2\alpha}{2v_0^2}x^2-(\tan\alpha)x-y_0=0
    [/tex]
    This leads to a complicated expression in x the positive root of which is the range.
    The problem now is finding
    [tex]
    \frac{dx}{d\alpha}
    [/tex]
    and setting it to zero to find the value of alpha to make x a maximum.
    John
     
  6. Aug 20, 2009 #5

    kuruman

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    Gold Member

    First off set y0 = 0. Then re-express all trig functions with sines and cosines. Then it should be obvious to you what's going on.
     
  7. Aug 20, 2009 #6
    With yo=0
    [tex]
    x=\frac{v_0^2}{g}\sin2\alpha
    [/tex]
    Then with
    [tex]
    \frac{dx}{d\alpha}=0
    [/tex]
    [tex]
    \frac{2v_0^2}{g}\cos2\alpha=0
    [/tex]
    This gives
    [tex]
    \alpha=\pi/4
    [/tex]
    However how do we solve this problem to include y0.
    The book answer to this question is:
    [tex]
    \sin\alpha=\frac{v_0}{\sqrt(2(v_0^2+gy_0))}
    [/tex]
     
  8. Aug 20, 2009 #7

    kuruman

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    Gold Member

    First you need to solve the equation you have with y=0
    [tex]

    0=y_0+(\tan\alpha)x-\frac{g(\sec^2\alpha)}{2(v_0)^2}x^2

    [/tex]

    for x. The larger of the two values is the range (Why?). Then, as tiny-tim said, maximize that function.
     
  9. Aug 20, 2009 #8

    tiny-tim

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    Science Advisor
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    Hi kuruman! :smile:
    I think you meant set y = 0. :wink:

    But using sec2 = tan2 + 1 is probably easier. :smile:
     
  10. Aug 21, 2009 #9
    I have set y=0
    Then the equation is:
    [tex]
    \frac{g\sec^2\alpha}{2v_0^2}x^2-(\tan\alpha)x-y_0=0
    [/tex]
    Now need to express x explicitly in terms of alpha, then find dx/d(alpha)
    then equate
    [tex]
    \frac{dx}{d\alpha}=0
    [/tex]
    The equation is a quadratic in x and the roots are complicated and finding dx/d(alpha) is the problem?
    John
     
  11. Aug 21, 2009 #10

    tiny-tim

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    … and the day is long :redface:, and the wages are small :redface:

    get on with it!! :rolleyes:
     
  12. Aug 21, 2009 #11
    Y0 is the initial distance? from what?

    what is y0
     
  13. Aug 21, 2009 #12
    [tex]
    y_0
    [/tex]
    is the height above ground level from which the projectile is fired.
    John
     
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