Help - Physics problem spiraling out of control!

Factual Questions
1

Have you ever tried to do a math problem and it quickly gets out of hand with no answer in sight? This is one of those. It probably didn’t help that I tried to start it so late at night after having several beers, but what the hey?

Anyway, trying to solve this puzzle:

Given a projectile launched from a flat plane at an angle theta from the horizontal with a velocity v (neglect air resistance), what is the maximum horizontal distance it can travel and what angle does it need to be shot at?

Simple, right? I thought so when I started, but I just keep going around in circles. I can’t believe that I have forgotten how to do this stuff!

BTW, the real problem I’m trying to solve (from the Spring 2001 “The Bent”) is what is the total volume of sky that can be reached by a shell from an anti-aircraft gun on a flat plane neglecting air resistance? i.e. what is the envelope of coverage? The gun can be shot at any azimuth or elevation by rotation around it horizontal and vertical axis. If you have any insight into this problem I’d appreciate it also!

Thanks!

2

To answer the first question, what launch angle w gives maximum range, we must find the value of w at which x[sub]f[/sub] is maximised.

If no air resistance, then only force is gravity and acceleration is –g acting vertically.
Then:
dy/dt = v[sub]y/sub = v[sub]y,0[/sub] - gt
And
dx/dt = v[sub]x/sub = v[sub]x,0[/sub]

So dy = (v[sub]y,0[/sub] - gt)dt and y =v[sub]y,0[/sub]t - 1/2gt[sup]2[/sup]
dx=v[sub]x,0[/sub]dt and x = v[sub]x,0[/sub]t

x[sub]f[/sub] = x at which y = 0:
y =v[sub]y,0[/sub]t - 1/2gt[sup]2[/sup]=0
Solving for t we get t = 0 or 2v[sub]y,0[/sub]/g
Taking the second solution and substituting for t:
x[sub]f[/sub]=2v[sub]x,0[/sub]v[sub]y,0[/sub]/g
=2v[sub]0[/sub]Cos(w)v[sub]0[/sub]Sin(w)/g
=2v[sup]2[/sup]/g Cos(w)Sin(w)
=v[sup]2[/sup]/g Sin(2w)

Now find at what angle w x[sub]f[/sub] is maximised by differentiating:
d(x[sub]f[/sub])/dw=0
(2v[sub]0[/sub][sup]2[/sup]/g)Cos(2w)=0
Cos(2w)=0
2w=90 or 270 degrees
Taking first solution w = 45 degrees.

Therefore the maximum range is achieved at a launch angle of 45 degrees.
This range will be v[sub]0[/sub][sup]2[/sup]/g.

3

Now I remember why I hated physics…

4

I like string.

5

Hey, I went ahead and did the problem you mentioned, if you’d like to use my value to check yours, and vice versa.

Firstly, I agree with what hibernicus said, but I’m not going to go into as much depth. I stared with the equations for vertical and horizontal velocity and integrated them to get the equations for vertical and horizontal position:

z = v[sub]0[/sub]t×sin([Sym]q[/Sym]) - gt[Sup]2[/Sup]/2
r = v[sub]0[/sub]t×cos([Sym]q[/Sym])

For a given horizontal distance r and elevation angle [Sym]q[/Sym], we can determine how high the projectile will be by eliminating the parameter t:

z = r×tan([Sym]q[/Sym]) - gr[sup]2[/sup]×sec[sup]2/sup / 2v[sub]0[/sub][sup]2[/sup]

This function z([Sym]q[/Sym]) has a maximum value of

z[sub]max[/sub] = v[sub]0[/sub][sup]2[/sup] / 2g - (g / 2v[sub]0[/sub][sup]2[/sup])r[Sup]2[/Sup]

when

tan([Sym]q[/Sym]) = v[sub]0[/sub][sup]2[/sup] / gr

Note that the largest value of r for which z[sub]max[/sub] [Sym]³[/Sym] 0 is

r = v[sub]0[/sub][sup]2[/sup] / g

This is the maximum distance you can shoot, which (fortunately for me) is the same value hibernicus got. I’ll call this value R to simplify things. Then we also have

z[sub]max[/sub] = R / 2 - r[Sup]2[/Sup] / 2R

Now we know what kind of envelope we’re dealing with. The cylindrical equation z = z[sub]max[/sub] describes a paraboloidal bowl, opening down. We can determine its volume with a volume integral over rdzdrd[sym]f[/sym]. The ranges are:

r = 0…R
z = 0…z[sub]max[/sub]
[sym]f[/sym] = 0…2[sym]p[/sym]

And the integral evaluates to

V = ([sym]p[/sym] / 4)R[sup]3[/sup], where R is how I defined it earlier. Is that what you’re getting?

6

You may or may not wish to also mention the line directly above the gun, where they might not wish to fire at for fear of missing. Of course, it won’t affect your volume measurements if you assume the gun is of inifinitesimal caliber.