Shooting a .45 caliber hand gun on the moon "launches" you a mile? My co-worker and I were having a discussion about an activity that he did while at a teachers math conference. The activities was to rate the importance of objects. There were 15 objects and you are stranded on the moon 200 miles away from your base. One of the objects was a .45 caliber handgun with 2 rounds of ammunition. I put this as low importance (along with the matches lol) because there's nothing that's after you and it's not going to "launch" you any significant amount of distance. My co-worker had said that the guy running the activities put it higher because it can launch you a mile or two. I am a sophomore civil engineering student and I'm in a dynamics class right now so I did my own basic calculations to see how far you could go. What I found on the internet is a .45 caliber bullet fired from a handgun has anywhere from 500-1000 joules of energy. I am not sure if the expelled gasses would make a significant difference. For my calculations I gave the bullet 1500 joules just to see what happens and to try to account for the expelled gasses. What I got is Newton's third law is every action has a equal and opposite reaction. 1500 joules in the bullet means 1500 joules transferred to the man firing it. Let's say the man is 200 lb on earth so about 90kg. 1500 joules / 90kg = 16.666666 m/s after shot Let's say he fires it at 45 degrees towards the ground for max distance. Vxi=16.66666cos(45) Vyi=16.66666sin(45) Kinematics equation for constant acceleration Vf=Vi+at Assume it's relatively flat and moons gravity is 1/6 of earths Vyf=-16.66666sin(45) -16.66666sin(45)=16.66666sin(45)+(-9.81/6)t t=14.416 Distance in x direction is velocity times time Distance=16.66666cos(45)*14.416 Distance is about 170 meters making the bullet fairly insignificant as your 200 miles from base and you have a lot of gear with you. If there's anything I'm not accounting for or if I messed up somewhere let me know.