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working horizontally: time = dist/speed = 7500/3000 = 2.5

working vertically: distance= initial velocity*time + 0.5* accelleration*time^2

distance=0*2.5+0.5*-9.8*2.5^2 = -61.25 feet

it drops by 61 and 1/4 feet

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Q: A bullet travels at 3000 FPS. If target is 7500 feet away how far does the bullet drop from line of sight?
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