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Thread: [SOLVED] Prove max distance angle of shooting a cannon

  1. #1
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    [SOLVED] Prove max distance angle of shooting a cannon

    In class we proved that the best angle (ignoring air resistance) for shooting a cannon at a height of 0 is $\displaystyle 45 deg$.

    How would i prove what the best angle (less than or greater than 45) is if the cannon is raised to some Height, $\displaystyle H$, where $\displaystyle H>0$. I dont need the specific angle i just need to prove it is less than or greater than 45 degrees. I think its less than but I am not sure.

    Here are the initial values of everything.
    V=velcoity in m/s that the ball is shot at
    $\displaystyle \theta$=angle (which we are trying to optimize)
    H=starting height (greater than 0)
    X=distance in terms of t
    Y=Height in terms of t

    $\displaystyle X= Vcos(\theta)t$
    $\displaystyle Y= -4.9t^2 + Vsin(\theta)t + H$

    From here i would normally find where $\displaystyle Y=0$.
    So i used the quadratic equationa and got:

    $\displaystyle \frac
    {-fsin(\theta)t + \sqrt{f^2 sin^2(\theta) t^2 - 9.8t^2 h}}{-9.8t^2}
    $
    or the minus solution.

    From here i am not sure what the next step would be to solve for t or even if this last step helps.
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  2. #2
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  3. #3
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    $\displaystyle = 0$


    $\displaystyle
    -fsin(\theta)t + \sqrt{f^2sin^2(\theta)t^2 - 9.8t^2h} = 0
    $

    $\displaystyle
    f^2 sin^2(\theta)t^2 + f^2sin^2(\theta)t^2-9.8t^2h = 0
    $

    $\displaystyle
    2f^2sin^2(\theta)t^2 - 9.8t^2h = 0
    $

    factor out $\displaystyle 2t^2$:
    $\displaystyle
    2t^2(f^2sin^2(\theta)-4.9h)=0
    $

    t=0 or....here is where i end
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  4. #4
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    I looked so hard for a solution...Thanks!
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