[SOLVED] Prove max distance angle of shooting a cannon

**snaes** · September 24th 2009, 10:11 AM

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 {-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.

**skeeter** · September 24th 2009, 10:16 AM

http://www.usna.edu/Users/physics/mu...Projectile.pdf

**snaes** · September 24th 2009, 10:22 AM

$= 0$

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

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

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

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

t=0 or....here is where i end

**snaes** · September 24th 2009, 10:24 AM

I looked so hard for a solution...Thanks!

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[SOLVED] Prove max distance angle of shooting a cannon

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