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Homework Help: Optimal Angle

  1. Aug 18, 2005 #1
    I shoot a cannon from a cliff of height [itex] h [/itex], with an initial velocity [itex] v_0 [/itex] and angle of elevation [tex] \theta [/tex]. If [itex] h > 0 [/itex], what [tex] \theta [/tex] will maximize the cannon's range (how far the cannonball lands from base of the cliff), assuming no air resistance?
    The cannonball's height at any moment within its trajectory is
    [tex] y = h + v_0 t\sin \theta - \frac{{gt^2 }}{2} [/tex]

    The cannon will land at ground zero, so I find that the time spent in the air is:
    [tex] y = 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} }}{g} [/tex]

    And therefore the range is:
    [tex] x = v_0 \frac{{v_0 \sin \theta + \sqrt {v_0^2 \sin ^2 \theta + 2gh} }}{g} \cos \theta = \frac{{v_0 }}{g}\left( {v_0 \sin \theta \cos \theta + \left( {\cos \theta } \right)\sqrt {v_0^2 \sin ^2 \theta + 2gh} } \right) [/tex]

    *But, which [tex] \theta [/tex] will maximize the range? Here, I simply find the roots of [itex] {dx}/{d\theta} [/itex]:
    [tex] \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 [/tex]
    [tex] 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 [/tex]

    To simplify this equation, I can multiply both sides by
    [tex] \frac{{\sqrt {v_0^2 \sin ^2 \theta + 2gh} }}{{v_0 \sin \theta }} [/tex]
    and continue to get :smile::

    [tex] \left( {\cos 2\theta } \right)\sqrt {v_0^2 + 2gh\csc ^2 \theta } \; + \; v_0 \cos 2\theta \; + \; 2v_0^{ - 1} gh = 0 \Rightarrow - \sqrt {v_0^2 + 2gh\csc ^2 \theta } = v_0 \; + \; 2v_0^{ - 1} gh\sec 2\theta \Rightarrow [/tex]

    [tex] \csc ^2 \theta = 2\sec 2\theta \left( {1 + v_0^{ - 2} gh\sec 2\theta } \right) [/tex]

    Now, we can multiply both sides by [itex] \cos 2\theta [/itex], to get
    [tex] \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 ?? [/tex]
    And now I'm stuck; any ideas?
    Last edited: Aug 18, 2005
  2. jcsd
  3. Aug 18, 2005 #2


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    I haven't worked it all out, but it seems like this should lead directly to the solution: Take the equation while it still says
    [tex] \csc ^2 \theta = 2\sec 2\theta \left( {1 + v_0^{ - 2} gh\sec 2\theta } \right) [/tex]
    Now raise both sides to the -1 power:
    [tex]\sin^2\theta = \frac{\cos 2\theta} {2(1+ v_0^{ - 2} gh\sec 2\theta) } = \frac{\cos^2 2\theta} {2\cos 2\theta+ 2v_0^{ - 2} gh}[/tex]
    Then substitute for [itex]\sin^2 \theta[/itex]:
    [tex]\sin^2\theta = 1 - \cos^2 \theta = 1 - \frac{\cos 2\theta +1} {2}[/tex]
    Now all of the trig functions are [itex]\cos 2\theta[/itex] so its just a quadratic equation.
  4. Aug 18, 2005 #3
    [tex] 1 - \cos 2\theta = \frac{{\cos ^2 2\theta }}{{v_0^{ - 2} gh + \cos 2\theta }} \Rightarrow \left( {1 - \cos 2\theta } \right)\left( {v_0^{ - 2} gh + \cos 2\theta } \right) = \cos ^2 2\theta \Rightarrow [/tex]
    [tex] v_0^{ - 2} gh \; + \; \left( {1 - v_0^{ - 2} gh} \right)\cos 2\theta \; - \; 2\cos ^2 2\theta = 0 \Rightarrow 2v_0^2 \cos ^2 2\theta \; - \; \left( {v_0^2 - gh} \right)\cos 2\theta \; - \; gh = 0 \Rightarrow [/tex]

    [tex] \cos 2\theta = \frac{{v_0^2 - gh + \sqrt {v_0^4 + 6v_0^2 gh + g^2 h^2 } }}{{4v_0^2 }} \Rightarrow \theta = \frac{1}{2}\cos ^{ - 1} \left[ {\frac{1}{4} - \frac{{gh - \sqrt {\left( {v_0^2 + gh} \right)^2 + 4v_0^2 gh} }}{{4v_0^2 }}} \right] [/tex]

    *Is this correct??
    *Can the expression within the arccosine be further simplified?
    Last edited: Aug 18, 2005
  5. Aug 19, 2005 #4


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    You get your last expresssion here by dividing by [tex]cos2\theta[/tex], but when h = 0, [tex]Cos2\theta = 0[/tex].

    When projecting upon horizontal ground, the optimum angle is 45ยบ.

    In your final expression (last post), what value of [tex]\theta[/tex] do you get for h = 0?
  6. Aug 19, 2005 #5
    Well, the original problem explicitly stated
    Somewhat similar as to why [itex] a \ne 0 [/itex] when solving [itex] a x ^ 2 + b x + c = 0 [/itex] for [itex] x [/itex]. You cannot divide by [itex] 2 a [/itex] in the quadratic formula if [itex] a = 0 [/itex]. However, if [itex] a = 0 [/itex], then there are simpler ways of solving quadratics and get solutions for x. But why is this an issue here? What's wrong with it, if I consider only [itex] h > 0 [/itex] ?

    However, we can substitute [itex] h = 10 ^ {-10} [/itex]...or any number sufficiently close to zero--->we can also substitute h = 0. Also, let [tex] v_0 = 10 [/tex]. Needless to say, [itex] g = 9.8 [/itex].
    Therefore, the optimal angle would be:
    [tex] \theta = \frac{1}{2}\cos ^{ - 1} \frac{1}{2} = \frac{\pi }{6} [/tex]

    However, the correct answer is [tex] \theta = {\pi/4} [/tex].

    *Thus, I think my formula for the optimal [tex] \theta [/tex] for maximum range is incorrect........But why? Where have I gone wrong? Seriously, why is it wrong do divide by [itex] \cos {2\theta} [/itex]...I mean, I did state [itex] h > 0 [/itex].
    |*It seems there's a problem. But what exactly is it? How can it be solved?
    And what is the REAL, the True formula to calculate the [tex] \theta [/tex] ?

    *According to a friend's calculator, the optimal [tex] \theta [/tex] for maximum range is given by the formula:

    [tex] \theta = \cos ^{ - 1} \left[ {\frac{{\sqrt {2\left( {2gh + v_0^4 } \right)} }}{{2\sqrt {gh + v_0^4 } }}} \right] [/tex]

    But is this truly the formula to find the optimal [tex] \theta [/tex] for the maximum range? If so, how is it derived?
    Last edited: Aug 19, 2005
  7. Aug 19, 2005 #6


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    In the quadratic formula, you can't have a=0, not only because division by zero is out, but because if a = 0, then you don't have a quadratic, just a linear expression in x. The QF is for quadrtic expressions, so having a != 0 is a condition of using it.

    In your optimum angle problem, why do you not allow h=0?
    h=0 is the state of the problem when projecting upon horizontal ground and any (general) expression should be able to take h=0 as one possible value.
    Also, putting h=0 allows you to check the accuracy of your own general solution for one specific case.

    I solved this problem in another forum, some time ago, but I can't find my solution. I'll have another look and see if I can figure out something.
    Last edited: Aug 19, 2005
  8. Aug 19, 2005 #7
    From reading your previous post, it seemed that you saw h = 0 as a problem...a problem when dividing by [itex] cos {2\theta} [/itex] when solving [itex] {dx}/{d\theta} = 0 [/itex] for for [tex] \theta [/tex]. Would it be this step that invalidates my original formula for the [tex] \theta [/tex] ?
    In some other forum (non-PF)...or just in another section of PF?
    Can you remember some phrases/quotes, that I may google-search for it?

    Aside from that, how fares the calculator solution?
    [tex] \theta = \cos ^{ - 1} \left[ {\frac{{\sqrt {2\left( {2gh + v_0^4 } \right)} }}{{2\sqrt {gh + v_0^4 } }}} \right] [/tex]
    Not sure how this was derived...but it appears to work:
    @h=0, [itex] \theta = {\pi}/{4} [/itex], and [itex] {d\theta}/{dh} < 0 [/itex] as expected.
  9. Aug 19, 2005 #8


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    This can be a problem, but as a problem can be circumvented by taking out [tex]cos2\theta[/tex] as a factor, which can be equated it to zero as a solution of the expression. This would give you [tex]\theta = \pi/4[/tex] as one solution! So now I don't think that's the problem.
    It's from TSR maths forum, but I've already searched for it with no luck. I'm sorry but I can't remember any significant details - but it was the exact same problem. Fiinding the optimum angle for shooting off a cliff/hill top.
    Still working on it. But there's a problem in the expression you gave. gh and [tex]v_0^4 [/tex] should have the same units. Shouldn't it be [tex]v_0^2[/tex]
  10. Aug 19, 2005 #9


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    Worked throught the problem again, and as answer got,

    [tex]cos2\theta = \frac{gh}{gh + v_0^2}[/tex]

    which, in effect, is the same the one you got,

    [tex] \theta = \cos ^{ - 1} \left[ {\frac{{\sqrt {2\left( {2gh + v_0^4 } \right)} }}{{2\sqrt {gh + v_0^4 } }}} \right] [/tex]

    (once you chage the [tex]v_0^4[/tex] to [tex]v_0^2[/tex])

    Found your error - it was just a wrong sign!

    The final term [tex]+ \; 2v_0^{ - 1} gh[/tex] should be [tex]- \; 2v_0^{ - 1} gh[/tex]

    If you work through from there on, you should end up with the expression I got, which you can simplify to the one from your friend with the calculator.
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