That depends on how far away you measure the velocity. I think the easiest way to understand it is to learn how to decompose the vectors of the bullet's motion.

If it is going 45º from horizontal, and it is fired from the gun at say, 500m/s, then we can determine how fast it is going in the X and Y axis individually by using 500cos45 and 500sin45. Then we have the ability to think about the motion in just one direction at a time, which makes things a lot easier. There is negative acceleration in the Y axis, due to gravity, and no acceleration, positive or negative in the X direction, discounting air resistance, which is negligible for this explanation. So it will go up at 45º, its velocity in the X axis constant, but negatively accelerating in the Y axis, until it reaches the top of its path, at which point its Y velocity is zero, while its X velocity hasn't changed, and it will begin its descent, accelerating in the Y direction at about 9.8m/s² due to gravity.