# Canon shooting area in polar coordinates

1. Apr 8, 2014

### prehisto

1. The problem statement, all variables and given/known data
hi,guys.
The directions of shooting e=cos$\alpha$cos$\varphi$i+cos$\alpha$sin$\varphi$j+sin$\alpha$k
0<$\varphi$<=2π;$\varphi$ -horizontally
$\alpha$[0,π];$\alpha$ is vertically
initial speed=v0

I need to calculate the surface equation of canon shots (where it hits).
In other words equations of the surface which is made from canon hits.

2. Relevant equations

3. The attempt at a solution

My view is that i need to gain horizontal and vertical distance (r),and then i can get the desired equation.

So horizontal distance is r=t|vxy|=tv0|ex+ey|=tv0(cos$\alpha$cos$\varphi$i+cos$\alpha$sin$\varphi$j)1/2=tv0cos($\alpha$)

So now i need height or in other words the distance in vertical direction(z -direction) or all together?.. I have hard time recalling use of equations of motion.

Last edited by a moderator: Apr 8, 2014
2. Apr 8, 2014

### BvU

Is there a question in the problem statement ?
You'll find equations of motion here, or otherwise $\vec F = m\vec a$ is a good one too.

List what you need under 2 and use less shorthand than in

3. Apr 8, 2014

### prehisto

Im sorry,i edited the original post,tried to be more specific.

4. Apr 8, 2014

### BvU

Let me try to translate: you want to calculate where a cannon ball lands that is shot from a cannon that is placed on the x-y plane in an empty flatland with no atmosphere, and gravity acceleration g in the -z direction, as a function of $\phi, \theta$ v0 and g.
That right ? Something like this ?

Did you find something useful among the equations of motion? Or here or here ?

5. Apr 8, 2014

### prehisto

Yes,and the direction of cannon is described by $\vec{e}$=cos$\alpha$cos$\varphi$i+cos$\alpha$sin$\varphi$j+sin$\alpha$k

Because 0<$\varphi$<2pi and $\alpha$[0,pi].Result is not one spot where it would land,but a surface which im looking for.

Gravity g=-g$\vec{k}$

Yes,I found something useful,I think I could use dv=vvt-gt2/2

6. Apr 8, 2014

### BvU

Yes, so dv=0 has two solutions, one trivial at shooting off and one at "landing".
Bingo.