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Physics of a bullet

Kinetic Energy Released

The formula for kinetic energy is

KE = ½mass x velocity2

So for example, a small .45 caliber bullet weighing 15 grams and traveling at 288 meters per second yields is 619 joules of energy. Further explained, if a man weighing 139 lbs (63.2 kg) were to fall off of a bed, it would yield roughly the same energy as being shot by that bullet; the difference being with a fall the energy is dispersed through the entire surface area of the man's body versus a bullet where the focal point is a tiny circle.

KE = ½mass x velocity2
KE = (.015kg / 2) x (288 m/s x 288 m/s)
KE = 619 joules of energy

PE = mass x gravity x height
PE = 63.2kg x 9.81 m/s x 1 meter
PE = 619 joules of energy

So taking this information, let's plug in the numbers of the Apache's M230 automatic gun ammunition. We have each 30mm round weighing 350 grams and traveling at 800 met

Added: Apr-28-2009 
By: wilkinsonm
Iran, Arts and Entertainment
Tags: Physics, of, a-10, and, apache, bullets
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