Adult Children by Stephen Beals for April 06, 2017

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    Happy, happy, happy!!! Premium Member about 7 years ago

    …and a frog…

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    DM2860  about 7 years ago

    The point is that the gun hit the ground 4 feet later but the bullet hit the ground a few thousand feet away (1800 meters for a 9mm automatic or about a mile).

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    gigagrouch  about 7 years ago

    i thought a “dropped gun” was something found beside an unarmed suspect?

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    BubbleTape Premium Member about 7 years ago

    pretty sure that isn’t true. i’m no physicist but i’m thinking the force of the explosion on the bullet will be far greater than any force from the explosion on the gun – assuming the explosion happened after the gun had left the hand (even some grip from the hand would have an impact) — and then there is the issue of where the gun is pointed – if towards the ground, then he bullet will move quite quickly towards the ground, while the explosion would have pushed the gun in the opposite direction, slowing its decent even slightly; or if pointed straight or up – the bullet might not travel as far compared to being projected from a gun being held firm, but it will still travel further and longer than it would take the gun to hit the ground. Unless of course the person holding the gun is standing on top of a building, holding the gun over the edge when they dropped it because then it would a further distance to drop. That just adds in a whole bunch of unknown variables.

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    NCTom Premium Member about 7 years ago

    There is a whole lot of physics involved, Bubble is sort of right. If there was any upward angle of the gun above exact parallel with the ground, the bullet will gain elevation before dropping and the gun would be pushed slightly in a downward direction also. But even if the gun is pointed exactly parallel to the earth’s surface at the time of firing and dropping the gun, the bullet would hit the ground slightly after the gun, because as the bullet traveled away from the gun, the earth’s surface would curve away from the exact parallel vector of the bullet at the start. Of course, we are assuming a perfectly flat earth’s surface also. But an interesting idea!

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