(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

A gun can fire shells in any direction with the same speed [tex]v_{0}[/tex]. Ignoring air resistance and using cylindrical polar coordinates with the gun at the origin and [tex]z[/tex] measured vertically up, show that the gun can hit any object inside the surface

[tex]z = \frac{v^{2}_{0}}{2g} - \frac{g}{2v^{2}_{0}}\rho[/tex]

Describe this surface and comment on its dimensions.

2. Relevant equations

[tex]m\ddot{z} = mg[/tex]

[tex]\ddot{\rho} = 0[/tex]

(Angular acceleration is unimportant because the problem has radial symmetry)

3. The attempt at a solution

The surface is a sort of parabolic dome (circular paraboloid, I believe would be the technical term?) with its vertex at the maximum height the projectile would go if fired straight up. This is fairly obvious from the given equation.

Just curious if I could get the solution without using cylindrical coordinates explicitly, I attacked the problem naturally by assigning an angle [tex]\theta[/tex] between the horizontal and the initial velocity of the projectile. Then the equations fall out as follows:

[tex]v_{0}sin(\theta) = v_{z0}[/tex]

[tex]z = \frac{v^{2}_{0}sin^{2}(\theta)}{2g}[/tex]

[tex]\rho = v_{0}cos(\theta)t[/tex]

[tex]t = \frac{v_{0}sin(\theta)}{g}[/tex]

[tex]\rho = \frac{v^{2}_{0}cos(\theta)sin(\theta)}{g}[/tex]

I then made the substitution into the equation for z.

[tex]z = \rho^{2}(\frac{g}{2v^{2}_{0}cos^{2}(\theta)})[/tex]

(LaTeX is not displaying this equation for me - it should read:

z = rho^2*(g/(2v_0^2*cos^2(theta))

Obviously, the cosine term is not in agreement with the given equation. I may be missing some kind of identity or making some kind of mistake, but I figured I'd move on to trying explicit cylindrical coordinates (since these were kind of fudged) and seeing if the problem resolved itself.

Integrating the differential equations once we have

[tex]\dot{z} = -gt + \dot{z_{0}}[/tex]

[tex]\dot{\rho}= \dot{\rho_{0}}[/tex]

Integrating the z equation once more we have

[tex]z = -\frac{1}{2}gt^{2}+\dot{z_{0}}t[/tex]

I noticed that we can substitute in [tex]t = \frac{\rho}{\dot{\rho_{0}}}[/tex]

[tex]z = -\frac{1}{2}g(\frac{\rho}{\dot{\rho_{0}}})^{2}+\dot{z_{0}}(\frac{\rho}{\dot{\rho_{0}}})[/tex]

Furthermore, we can make the substitution [tex]\dot{\rho_{0}} = \sqrt{v^{2}_{0}-\dot{z_{0}}^{2}}[/tex]

[tex]z = \frac{\dot{z_{0}}\rho\sqrt{v^{2}_{0}-\dot{z^{2}_{0}}-\rho^{2}g}{2(v^{2}_{0}-\dot{z_{0}}^{2}})[/tex]

(These two equations also aren't showing up. They should read

dotted rho_0 = sqrt(v_0^2 - dotted z_0^2)

z = (dot z_0 * rho * sqrt(v_0^2 - dot z_0^2) - rho^2 * g)/(2*(v_0^2 - dot z_0^2)

respectively)

Here is where I get stuck, because I don't see how to resolve the problem of having either

[tex]\dot{z_{0}}[/tex] or [tex]\dot{\rho_{0}}[/tex]

appear in the equation. Furthermore, on the whole, it seems like I've actually moved further away from the given solution rather than closer to it.

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# Homework Help: Simple Projectile in Cylindrical Coordinates

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