What would the field of view be on the phantom 3 pro Camera flying at 500 Meter AGL max altitude and the camera pointing straight down.

The 94° FOV is measured diagonally If my calculations are correct, at 500 m the camera would cover an area measuring 900 m x 675 m

Why would you use 16:9 it's just 4:3 with a bit cut off the top and bottom. Wouldn't you use the biggest area and crop yourself if required? @500m - 4:3 gives you 900m x 675m @500m - 16:9 gives you 900m x 506m

Here's the math... Consider that 94-deg is a cone with the camera at the apex. To compute the base you need to divide the cone in half to result in a right triangle so to be able to apply the trigonometric function of "B(base)/H=tan-1 ("H" being the height, or altitude). But since we're solving for "B" we need to use the tangent of 47-deg (half of 94-deg), which is 1.0724. At a height of 500m x 1.074 (the tangent) B=536.18m. Remember that this half the cone so we need to double this to get back to the 94-deg FOV, which rounded is 1072m. This is the FOV at 500m per your question. I'm OK with geometery, but what you will actually see in your final image will be determined by your format and that is for someone else to compute.

All I know is that I shoot in 16:9 and at more modest heights of around 300', objects may appear closer than they really are. It's taking some getting used to because with my Q500 the FOV is 115- much wider. With 94, when I look back at my footage I'm always surprised that I wasn't as far out (close to) areas that I thought I was, looking at my video stream why flying. I suppose I have to really get a grip on it by full down facing tilt to see where the bird really is, in relation to where I think it actually is. This has caught me off guard once setting up a POI in a bit of a rush and thinking I was very much close enough to the target but realizing I really wasn't when I started the orbit with a not so large radius. Something to get used to, for me.