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Deriving projectile motion result

  1. Oct 29, 2014 #1
    1. The problem statement, all variables and given/known data
    A gun can fire an artillery shell with a speed v_0 in any direction. Show that a shell can strike any target within the surface given by:

    (gr)^2 = (v_0)^4 - (2gz)*(v_0)^2

    where z is height and r is horizontal distance the target is from the gun.

    2. Relevant equations
    Some useful equations that trivially derived:

    (v_0*cos(k))^2 = v_0^2 -2g*z_max

    z_max = ((v_0*sin(k))^2) / 2g.


    3. The attempt at a solution
    I'm having problems getting anywhere near the answer. I tried playing around with the first term after substituting (r / (r^2+z^2)^1/2) for cos(k) ( can you even do this? the surface is an arc.)

    Anyone want to help guide me here?
     
  2. jcsd
  3. Oct 29, 2014 #2

    ehild

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    You need the equation of the trajectory of the shell z=f(r), not only the maximum height and maximum distance. Check "projectile motion" in your lecture notes.
     
  4. Oct 29, 2014 #3
    looks like I have an equation for that.
    It can be trivially shown that:

    z = tan(k)*x - (g / (2(vcos(k))^2)*x^2.

    Do I use tan(k) = z/r
    and
    cos(k) = r/(r^2+z^2)^1/2
    ?
     
  5. Oct 29, 2014 #4

    PeroK

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    I'd start by drawing a diagram of this surface. Note that it reduces to z as a quadratic in r, so should be easy to draw.

    Then you might see a simple approach to solving it.
     
  6. Oct 29, 2014 #5

    ehild

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    k is the angle the initial velocity v makes with the horizontal. x and z or r and z are the horizontal and vertical coordinates of the target which has to be a point of the trajectory.
     
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