Canon shooting area in polar coordinates

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Homework Help Overview

The discussion revolves around calculating the surface equation of cannon shots in polar coordinates, specifically focusing on the trajectory of a cannonball shot from a cannon placed on the x-y plane. The original poster seeks to understand how to derive the equation representing the surface formed by the cannon hits, considering factors such as initial speed and angles of projection.

Discussion Character

  • Exploratory, Conceptual clarification, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the need to calculate both horizontal and vertical distances to derive the desired surface equation. There are questions about the clarity of the problem statement and the use of equations of motion. Some participants suggest listing necessary components and using less shorthand in the original post.

Discussion Status

The discussion is ongoing, with participants attempting to clarify the problem and explore the necessary equations. Some guidance has been offered regarding the equations of motion, and there is an acknowledgment of the complexity of deriving a surface rather than a single landing spot.

Contextual Notes

There are indications of missing information and the need for clearer definitions of variables and conditions, such as the effects of gravity and the specific setup of the cannon's firing angles.

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Homework Statement


hi,guys.
The directions of shooting e=cos[itex]\alpha[/itex]cos[itex]\varphi[/itex]i+cos[itex]\alpha[/itex]sin[itex]\varphi[/itex]j+sin[itex]\alpha[/itex]k
0<[itex]\varphi[/itex]<=2π;[itex]\varphi[/itex] -horizontally
[itex]\alpha[/itex][0,π];[itex]\alpha[/itex] 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[itex]\alpha[/itex]cos[itex]\varphi[/itex]i+cos[itex]\alpha[/itex]sin[itex]\varphi[/itex]j)1/2=tv0cos([itex]\alpha[/itex])

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[itex]\alpha[/itex]cos[itex]\varphi[/itex]i+cos[itex]\alpha[/itex]sin[itex]\varphi[/itex]j)1/2=tv0cos([itex]\alpha[/itex])
 
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 [itex]\vec{e}[/itex]=cos[itex]\alpha[/itex]cos[itex]\varphi[/itex]i+cos[itex]\alpha[/itex]sin[itex]\varphi[/itex]j+sin[itex]\alpha[/itex]k

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

Gravity g=-g[itex]\vec{k}[/itex]

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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