Having this discussion on another board, basically stemmed from one member saying that a bullet would be in flight for "over 5 seconds"in response to something else someone said about a 0.50 caliber sniper rifle firing at a target 2 miles away. I figured that negating air resistance, the time would be 3.857 seconds (using Vf^2 = Vi^2 + 2ad then putting finding average velocity (Vf + Vi)/2 and substituting that into v = d/t) "d" is two miles, or 10,560 feet, "Vi" is 2,800 feet per second, and "a" is -32.15 feet per second. So assuming that was right, I found that for the time to be "over 5 seconds" (I used 5.5), the final velocity would have to be 1,040 feet per second (using v = d/t, solving for v as average velocity, multiplying Va by two and subtracting initial velocity to find the new Vf) is that correct? Anyway, none of that really matters, just background, my question is, how do you take wind resistance into account in kinematics, and how high would the wind speed have to be to cause a 63% loss of velocity? I'd preferably like to do this without calculus. Thanks in advance.