Dale

Brainkrieg,
Here is how to figure out (roughly) how far a crossbow should shoot, if you are shooting at zero degrees (that is, straight ahead, no elevation at all).
First off, an object falling near the earth's surface, is accelerated downward at 32.15 feet/second/second (9.8 meters/second/second). It does not matter if it is pointed straight up, straight down, or anywhere in between. If it is pointing upward of horizontal, the acceleration of gravity will pull the object down until it is no longer moving upward, and thenceforth it will move downward faster and faster. But the crossbow bolt is being shot horizontally, so it is at rest as far as its vertical motion is concerned.
So. The distance an object moves under a constant acceleration is:
x = ^{1}/_{2} a t^{2}
The above formula assumes the object is initially at rest; there's a couple of terms I left out that deal with the initial position and velocity, if it is already moving.
If we plug in 9.8 m/s^{2} for acceleration "a", we see that after one second the bolt will have fallen 4.9 meters (16.1 feet).
Let us rearrange the formula and find out how long it will take the bolt to fall 5 feet, or 1.5 meters (assuming the crossbow is held at shoulder level and you are not extremely tall).
t = sqrt(2x/a)
Note: "sqrt" means "square root". There's a standard symbol for it, but the computer character set used here does not include that symbol. OK, where were we? Oh, yeah, now substitute in the values for distance "x" and acceleration "a":
t = sqrt(2 * 1.5 / 9.8 )
t = 0.55 second
Now, how fast is the bolt moving horizontally? Let's assume it's going 100 ft/sec (30.5 m/s). Ignoring air resistance, which is probably safe to do IN THIS CASE (because it's only flying for half a second), the bolt will travel 30.5 / 0.55 = 55.5 meters (182 feet) before it hits the ground.
Thus ends today's lesson in classical mechanics (basic physics).
