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Projectile motion - surface of max range

  1. Feb 4, 2009 #1
    Doing an Analytical Mechanics course and we're just reviewing some classical physics and this question came up. Cheers for any help.

    1. The problem statement, all variables and given/known data

    A gun can fire shells in any direction with the same speed v. Ignoring air resistance and using cylindrical polar coordinates with the gun at the origin and z measured vertically up, show that the gun can hit any object inside the surface:

    z = v²/(2g) - ρ²g/(2v²)

    (where ρ & z are as per cylindrical coordinates)


    3. The attempt at a solution

    My thought was that the surface is the set of points which are the farthest the gun can reach. As such the max vertical height can be found by conservation of energy:

    Max height when all vertical component of KE turned into PE.

    i.e. if θ is the angle trajectory from the ground:

    1/2 m (vsinθ)² = mgz

    Rearranging:

    z = (vsinθ)²/(2g) , which I think gives the max height the bullet will reach for a given θ. (Note that if θ = 90° => ρ = 0 gives z = v²/(2g) as per the answer).

    Next I wanted to determine ρ at the max height. As the motion will be parabolic, the max height should be achieved at half of the range for a given θ.

    Invoking Newton's second law and bashing through algebra an equation for z in terms of ρ can be found (by eliminating θ), and then setting z = 0 a solution for ρ(range) can be found, which when halved is ρ(max z):

    ρ(max z) = 2v²/g sinθcosθ

    Where I'm at

    At this stage I have tried bashing these two things around a bit with little success in deriving the given result. I feel the problem shouldn't be this complicated and that I may have taken a wrong turn early on.

    Any suggestions would be greatly appreciated.
     
  2. jcsd
  3. Feb 4, 2009 #2

    tiny-tim

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    Science Advisor
    Homework Helper

    Hi insynC! :smile:
    Yes … you're finding the range at ground level

    start again, and no short-cuts, this time! :wink:
     
  4. Feb 4, 2009 #3
    Thanks for the help, it came out.
     
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