[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.