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Math Help - [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 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, H, where 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
    \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

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

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

     \frac<br />
{-fsin(\theta)t + \sqrt{f^2 sin^2(\theta) t^2 - 9.8t^2 h}}{-9.8t^2}<br />
    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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    = 0


     <br />
-fsin(\theta)t + \sqrt{f^2sin^2(\theta)t^2 - 9.8t^2h} = 0<br />

     <br />
f^2 sin^2(\theta)t^2 + f^2sin^2(\theta)t^2-9.8t^2h = 0<br />

     <br />
2f^2sin^2(\theta)t^2 - 9.8t^2h = 0<br />

    factor out 2t^2:
     <br />
2t^2(f^2sin^2(\theta)-4.9h)=0<br />

    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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