[SOLVED] Prove max distance angle of shooting a cannon

• Sep 24th 2009, 10:11 AM
snaes
[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.
• Sep 24th 2009, 10:16 AM
skeeter
• Sep 24th 2009, 10:22 AM
snaes
http://www.mathhelpforum.com/math-he...7874729f-1.gif $\displaystyle = 0$

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

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

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

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

t=0 or....here is where i end
• Sep 24th 2009, 10:24 AM
snaes
I looked so hard for a solution...Thanks!