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Where's my Error?

  1. Aug 16, 2005 #1
    With no air resistance, I shoot a cannonball from a building of height [itex] h [/itex], with an initial velocity [itex] v_0 [/itex] at an angle of elevation [itex] \theta [/itex]. I'm trying to find the optimal angle for maximum range. I will use a parametric equation to represent the cannonball location at the time [itex] t [/itex]. So,
    *The cannonball's position is represented by the parametric function (well, obviously the initial position is at [itex] \left( {0,h} \right) [/itex]),
    [tex] \left\{ \begin{gathered}
    x = v_0 t\cos \theta \hfill \\
    y = h + v_0 t\sin \theta - \frac{{gt^2 }}{2} \hfill \\ \end{gathered} \right\} [/tex]
    and so time it takes for the cannonball to hit the ground is
    [tex] h + v_0 t\sin \theta - \frac{{gt^2 }}{2} = 0 \Rightarrow t = \frac{{ - v_0 \sin \theta + \sqrt {v_0^2 \sin ^2 \theta - 2gh} }}{{2h}} [/tex]
    *Thus, the range of the projectile is
    [tex] x = v_0 \cos \theta \frac{{\sqrt {v_0^2 \sin ^2 \theta - 2gh} - v_0 \sin \theta }}{{2h}} = \frac{{v_0 \cos \theta }}{{2h}}\left( {\sqrt {v_0^2 \sin ^2 \theta - 2gh} - v_0 \sin \theta } \right) [/tex]
    **But, which [tex] \theta [/tex] will maximize the range?
    [tex]\frac{{d}}{{d\theta }}\left[ {\frac{{v_0 \cos \theta }}{{2h}}\left( {\sqrt {v_0^2 \sin ^2 \theta - 2gh} - v_0 \sin \theta } \right)} \right] = 0 \Rightarrow [/tex]
    [tex] \frac{1}{{2h}} \cdot \left[ {v_0 \cos \theta \left( {\frac{{2v_0^2 \sin \theta \cos \theta }}{{2\sqrt {v_0^2 \sin ^2 \theta - 2gh} }} - v_0 \cos \theta } \right) - v_0 \sin \theta \left( {\sqrt {v_0^2 \sin ^2 \theta - 2gh} - v_0 \sin \theta } \right)} \right] = 0 \Rightarrow [/tex]
    *Well, I can factor out [itex] v_0 /2h [/itex], and set what remains equal to zero:
    [tex] {v_0 \cos ^2 \theta \left( {\frac{{v_0 \sin \theta }}{{\sqrt {v_0^2 \sin ^2 \theta - 2gh} }} - 1} \right) - \sin \theta \left( {\sqrt {v_0^2 \sin ^2 \theta - 2gh} - v_0 \sin \theta } \right)} = 0 \Rightarrow [/tex]
    [tex] \frac{{v_0^2 \sin \theta \cos ^2 \theta }}{{\sqrt {v_0^2 \sin ^2 \theta - 2gh} }} - v_0 \cos ^2 \theta + v_0 \sin ^2 \theta - \sin \theta \sqrt {v_0^2 \sin ^2 \theta - 2gh} = 0 \Rightarrow [/tex]
    [tex] v_0 \left( {\frac{{v_0 \sin \theta \cos ^2 \theta }}{{\sqrt {v_0^2 \sin ^2 \theta - 2gh} }} - \cos 2\theta } \right) = \sin \theta \sqrt {v_0^2 \sin ^2 \theta - 2gh} \Rightarrow [/tex]
    *Then I multiply both sides by [tex] {\sqrt {v_0^2 \sin ^2 \theta - 2gh} } [/tex] and then divide each side by [itex] v_0 [/itex]
    [tex] v_0 \sqrt {v_0^2 \sin ^2 \theta - 2gh} \left( {\frac{{v_0 \sin \theta \cos ^2 \theta }}{{\sqrt {v_0^2 \sin ^2 \theta - 2gh} }} - \cos 2\theta } \right) = \sin \theta \left( {v_0^2 \sin ^2 \theta - 2gh} \right) [/tex]
    [tex] v_0 \cos ^2 \theta - \left( {\cos 2\theta } \right)\sqrt {v_0^2 - 2gh\csc ^2 \theta } = v_0 \sin ^2 \theta - 2gh \Rightarrow [/tex]
    [tex] \begin{gathered}
    \sqrt {v_0^2 - 2gh\csc ^2 \theta } = 2gh + 1 \Rightarrow \hfill \\
    v_0^2 - 2gh\csc ^2 \theta = 4gh\left( {gh + 1} \right) + 1 \Rightarrow \hfill \\
    - 2gh\csc ^2 \theta = v_0^2 - 4gh\left( {gh + 1} \right) - 1 \Rightarrow \hfill \\
    \frac{1}{{\sin \theta }} = \sqrt {\frac{{v_0^2 - 4gh\left( {gh + 1} \right) - 1}}{{2gh}}} \Rightarrow \theta = \sin ^{ - 1} \left\{ {\left[ {\frac{{v_0^2 - 4gh\left( {gh + 1} \right) - 1}}{{2gh}}} \right]^{ - \frac{1}{2}} } \right\} \hfill \\ \end{gathered} [/tex]

    But I try a test with [itex] h = 50m,\;v_0 = 100\frac{m}{s} [/itex], and the optimal angle is not the one solved by the equation!
     
    Last edited: Aug 16, 2005
  2. jcsd
  3. Aug 16, 2005 #2
    Shouldn't that be +2gh rather than -2gh within the square root? I would rewrite the quadratic with a positive t^2 term.
     
    Last edited: Aug 16, 2005
  4. Aug 16, 2005 #3
    The denominator should be twice the t^2 term, which gives -g rather than 2h.
     
  5. Aug 16, 2005 #4
    Of course! The determinant must have +2gh and the denominator must have -g. Taking that into consideration:

    [tex] \begin{gathered}
    \left\{ \begin{gathered}
    x = v_0 t\cos \theta \hfill \\
    y = h + v_0 t\sin \theta - \frac{{gt^2 }}
    {2} \hfill \\
    \end{gathered} \right\} \Rightarrow \hfill \\
    y = 0,\;t = \frac{{v_0 \sin \theta + \sqrt {v_0^2 \sin ^2 \theta + 2gh} }}
    {g} \Rightarrow \hfill \\
    x = v_0 \cos \theta \frac{{v_0 \sin \theta + \sqrt {v_0^2 \sin ^2 \theta + 2gh} }}
    {g} = \frac{{v_0 }}
    {g}\left( {v_0 \sin \theta \cos \theta + \left( {\cos \theta } \right)\sqrt {v_0^2 \sin ^2 \theta + 2gh} } \right) \hfill \\ \end{gathered} \Rightarrow [/tex]
    [tex] \begin{gathered}
    \frac{{dx}}{{d\theta }} = \frac{{v_0 }}{g}\left[ {v_0 \cos 2\theta + \frac{{v_0^2 \sin \theta \cos ^2 \theta }}{{\sqrt {v_0^2 \sin ^2 \theta + 2gh} }} - \left( {\sin \theta } \right)\sqrt {v_0^2 \sin ^2 \theta + 2gh} } \right] = 0 \Rightarrow \hfill \\
    v_0 \cos 2\theta + \frac{{v_0^2 \sin \theta \cos ^2 \theta }}
    {{\sqrt {v_0^2 \sin ^2 \theta + 2gh} }} - \left( {\sin \theta } \right)\sqrt {v_0^2 \sin ^2 \theta + 2gh} = 0 \Rightarrow \hfill \\ \end{gathered} [/tex]

    *Then I multiply both sides of the equation by
    [tex] \frac{{\sqrt {v_0^2 \sin ^2 \theta + 2gh} }}{{v_0 \sin \theta }} [/tex]
    to get

    [tex] \begin{gathered}
    \left( {\cos 2\theta } \right)\sqrt {v_0^2 + 2gh\csc ^2 \theta } + v_0 \cos 2\theta + 2v_0^{ - 1} gh = 0 \Rightarrow \hfill \\
    - \sqrt {v_0^2 + 2gh\csc ^2 \theta } = v_0 + 2v_0^{ - 1} gh\sec 2\theta \Rightarrow \csc ^2 \theta = 2\sec 2\theta \left( {1 + v_0^{ - 2} gh\sec 2\theta } \right) \Rightarrow \hfill \\
    \cot ^2 \theta - 1 = 2\left( {1 + v_0^{ - 2} gh\sec 2\theta } \right) \Rightarrow \cot ^2 \theta - 2v_0^{ - 2} gh\sec 2\theta = 3 \Rightarrow ?? \hfill \\
    \end{gathered} [/tex]

    Hmm...and now I'm stuck! :redface: Any ideas?
     
  6. Aug 17, 2005 #5
    Please help :cry:...any ideas?
    How do I solve for [tex] \theta [/tex]?
    --such that
    [tex] \frac{{dx}}{{d\theta }} = 0 [/tex]
    ?

    Was there an error in my second attempt to calculate the [itex] \theta [/itex] such that
    [tex] \frac{{dx}}{{d\theta }} = 0 [/tex] :confused:
     
    Last edited: Aug 18, 2005
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