Thanks for the explanation Mitch. I respect you knowledge of math and physics, but the physics just sounds counterintuitive as it often does. If somebody fires 100' parallel to Earth, it will drop 2 ft for example. But if someone fired 100' from directly above, into the Earth below, there should be zero drop. So my contention is that if it were fired at an angle of 45 degrees, it should drop 1 ft. Where did I go wrong? Don't worry about it, let me give you an exact scenario:

JFK gets hit in the back 130' from the base of th SN (Z200) with a 400ft/sec bullet. I calculate the round flew 144'. What is its drop?

The absolute gravitational drop is the same no matter what direction you go. Even if you shoot straight down, the bullet still accelerates due to gravity, just in the direction of aim. But you're talking about drop below the point of aim. I wasn't sure if that's what you were getting at before. Anyway, let's go for a generalized equation covering the problem. For some angle A below horizontal and a horizontal distance D, the bullet travelling at velocity V, will travel a distance of D/cos(A) over a time span of D/(V*cos(A)). So the drop should be:

1/2 * G * (D/(V * cos(A)))^2.

G is 32.2 ~ 32 ft/s/s, so (1/2 * G) = 16

Now it's just about plugging in values. Using your D = 130' and assuming that the bottom of the SN window is 65', the angle A is tan-1(65/130) = 26.56, so we'll say 27deg. Distance traveled is easy to figure thanks to Pythagoras, so sqrt(65^2 + 130^2) = 145.3', so we agree on that part. Drop is going to be

16 * (130' / (400ft/s * cos(27)))^2

which boils down to:

2.12 feet worth of drop due to gravity

I'm curious as to the distance from the SN to the limo at frame 200. You have 130'. IIRC, the limo/SN distance has generally been held to be 195' at frame 220. The limo is rolling at about 12 mph at this point, that is a tad under 18ft/s. Frame 200 is 1.2 seconds before that, so the difference in distance should be around about 20', not 65'. Anyway, that's the back of the envelope version.