ORIGINAL: rwilson
Heard a guy on the radio (during the chaos of Arafat's funereal) talking about all the shooting up in the air and where all those bullets go and how dangerous they are. He said that according to the law of conservation of energy the bullet would be traveling the same speed when it hit the ground or whatever that it was traveling at when it left the muzzle. If this was true, a bullet would travel forever in a straight line, as long as nothing got in the way, right? Impossible, right? Say a bullet is fired perfectly straight up. It stops at the "top" and falls back to earth. At what speed will it be traveling when it strikes the ground?
The bullet fired vertically will often return base down, which limits its' terminal velocity even more.
IT DOES NOT RETURN WITH ITS' MUZZLE VELOCITY! According to Hatcher's Notebook, (In an entire chapter entitled BULLETS FROM THE SKY) returning bullets come down at unpredictable locations, and their return velocity cannot be accurately chronographed (obviously). However, impact dents made by returning 150-grain .30 caliber bullets which were fired vertically with a MV of 2700 FPS were about 1/16 of an inch in soft pine. These dents indicate that the velocity at impact was in the vicinity of 300 FPS.
While an 150 grain .308" bullet landing on you at 300 FPS could cause some damage and pain, 1/16" of an inch dent in soft pine is not indicative of a dangerous wound - it takes 1" of penetration in soft pine to equal a dangerous wound. BUT, a .50 BMG bullet of 718 grains would fall with close to 500 FPS, carrying around 400 Ft/Lb of energy - no joke!