L = R * m * vi * cos(90 – theta)

cos(90 – theta) is just sin(theta)

and R is the distance the projectile traveled, which is vi^2 * sin(2*theta) / g

so, we have: L = vi^2 * sin(2*theta) * m * vi * sin(theta) / g

We can combine the two vi terms and get:

L = vi^3 * m * sin(theta) * sin(2*theta) / g

What’s interesting is that angular momentum varies with the *cube* of the initial velocity. This is because, not only does increased velocity increase the translational momentum of the projectile, but it increase the *moment arm*, too. Also note that there might be a trig identity which lets you combine the two sin() terms, but nothing jumps out at me right at the moment.

Now, for the first part…

There are a few ways to attack this. Basically, you have to find the angle from the origin to the apogee (highest point) in the arc. Once we have that, we’ll know what angle the momentum vector makes with the moment-arm because, at the apogee, we know that all of the motion is *horizontal*.

Okay, so let’s get back to what we know:

L = d * m * v * cos(phi)

where d is the distance (length to the arm), m is mass, v is velocity, and phi is the angle the velocity vector makes with the arm. Let’s take these one by one…

m is still m.

v is going to be the *hoizontal* component of the initial velocity (all the vertical component got eliminated by the acceleration of gravity). So, v = vi * cos(theta)

d is going to be half of our distance R in part two (because, ignoring friction, the path of the projectile is a perfect parabola). So, d = vi^2 * sin(2*theta) / 2g

That leaves us with phi, the angle the horizontal velocity vector makes with the moment arm. To find *that*, we need to know what the angle from the origin to the apogee is. We can find *that* by taking the arc-tangent of the slope, if we know that. Well, we know the “run” part of the slope (it’s our “d” term), but not the rise.

The easy way to get the rise is by using conservation of energy. At the apogee, all of the *vertical* kinetic energy at the time of launch (1/2 * m * (vi * sin(theta))^2 ) has been turned into gravitational potential energy ( m * g * h ). Setting these equal, diving out the “m” and dividing “g” to the other side, we get:

h = 1/2 * (vi * sin(theta))^2 / g

So, there’s the rise. So, our *slope* is rise/run, so

slope = [ 1/2 * (vi * sin(theta))^2 / g ] / [ vi^2 * sin(2*theta) / g ]

The “g”s cancel. Astoundingly the “vi”s cancel, too. So, we get:

slope = [ 1/2 * sin(theta)^2 ] / [ sin(2*theta) ]

(It’s not too alarming that slope-at-apogee doesn’t depend upon vi, since that only determines the “magnitude” of the arc, but not it’s shape. Whether the overall flight of this thing is an inch or a mile, the arc “looks” the same).

Okay, so… using our double-angle trig identities, we know that sin(2*theta) = 2*sin(theta)*cos(theta), so…

slope = [ 1/2 * sin(theta)^2 ] / [ 2*sin(theta)*cos(theta) ] = tan(theta)/4

Okay, so the *angle* (which I’ll call “alpha”) that this slope makes with the x-axis is just: arctan(slope), so…

alpha = arctan( tan(theta) / 4 )

Alright… last bit. We need “phi”, the angle the (now-horizontal) momentum vector makes with that slope. Draw it on paper and you’ll see that phi = 180 – alpha

so, phi = 180 – arctan( tan(theta) / 4 )

Now, we go back to our original formula and plug it ALL in…

L = d * m * v * cos(phi)

becomes…

L = [ vi^2 * sin(2*theta) / 2g ] * m * [ vi * cos(theta) ] * [ cos( 180 – arctan( tan(theta) / 4 ) ) ]

Now, cos(180 – something) = cos(something), so we can simplify a little bit…

L = [ vi^2 * sin(2*theta) / 2g ] * m * [ vi * cos(theta) ] * [ cos( arctan( tan(theta) / 4 ) ) ]