Canon shooting area in polar coordinates

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Homework Statement


hi,guys.
The directions of shooting e=cos\alphacos\varphii+cos\alphasin\varphij+sin\alphak
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.

Homework Equations


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\alphacos\varphii+cos\alphasin\varphij)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.
some help,please?
 
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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
So horizontal distance is r=t|vxy|=tv0|e_x+e_y|=tv0(cos\alphacos\varphii+cos\alphasin\varphij)1/2=tv0cos(\alpha)
 
BvU said:
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

Im sorry,i edited the original post,tried to be more specific.
 
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 http://www.rabidgeek.net/physics-applets/projectile-motion/ ?

Did you find something useful among the equations of motion? Or here or here ?
 
BvU said:
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
Yes,and the direction of cannon is described by \vec{e}=cos\alphacos\varphii+cos\alphasin\varphij+sin\alphak

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

Gravity g=-g\vec{k}

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

Yes,I found something useful,I think I could use dv=vvt-gt2/2
 
Yes, so dv=0 has two solutions, one trivial at shooting off and one at "landing".
Bingo.
 
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