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Homework Help: Canon shooting area in polar coordinates

  1. Apr 8, 2014 #1
    1. The problem statement, all variables and given/known data
    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.

    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[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?
    Last edited by a moderator: Apr 8, 2014
  2. jcsd
  3. Apr 8, 2014 #2


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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
  4. Apr 8, 2014 #3
    Im sorry,i edited the original post,tried to be more specific.
  5. Apr 8, 2014 #4


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    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 ?
  6. Apr 8, 2014 #5
    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 im looking for.

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

    Yes,I found something useful,I think I could use dv=vvt-gt2/2
  7. Apr 8, 2014 #6


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