After a bullet leaves the barrel of the gun, it is no longer accelerating away from the gun, but instead beginning to drop in elevation due to the constant downward acceleration of gravity. If we consider air friction to be negligible, we can determine a bullet's trajectory by considering two separate components of that initial trajectory -- initial horizontal velocity (Vx) and initial vertical velocity (Vy) -- along with the angle to the ground at which the bullet was fired.

Determine horizontal velocity by using the equation: Vx = Vo Cos θ. If the launch angle was 45 degrees and initial velocity was 1500 meters per second, then: Vx = 1500 Cos 45 = 1500 (.0707) = 106 meters per second.

Determine vertical launch velocity by using the equation: Vy = Vo Sin θ. Using the same launch angle and initial velocity as before: Vy = 1500 Sin 45 = 1500 (.0707) = 106 meters per second.

Calculate how long it will be before the bullet stops rising (Vtop) -- which is when velocity in the vertical direction reaches zero -- given that gravity constantly pulls the bullet toward the ground (a). For this, we use the equation for velocity at a constant acceleration, Vtop = Vy +at, solved for time (t): t = (Vtop-Vy)/a. Since we know the acceleration of gravity to be -9.8 meters/ second^2, we can determine when the bullet in our example stops rising: t = (0-106)/-9.8 = 10.8 seconds.

Double the amount of time it takes for the bullet to reach peak altitude to find out the amount of time it takes for the bullet to strike the ground. In our example, the total flight time is: 10.8 x 2 = 21.6 seconds.

Determine how far along the ground the bullet will travel by using the formula: Distance (d) = Vxt + 0.5at^2. Considering an initial horizontal velocity of 106 meters per second, a flight time of 21.6 seconds, and a horizontal acceleration of zero, our bullet traveled: d = 106(21.6) + 0.5(0 x (21.6)^2)= 106(21.6) + 0 = 2289.6 meters along the ground.

#### Warning

Always follow all gun safety laws and protocols in your area.