The course or track of a projectile through the air after it leaves the gun is called the _trajectory_, and this has been studied both experimentally and theoretically, with interesting results. Assuming that the shot passed through empty space, or that the air offered no resistance to its passage, it would be very easy to trace the path of a projectile. Let us suppose that Fig. 80 represents a gun elevated at a high angle. The moment the projectile leaves the muzzle, gravity begins to act upon it, causing it to move vertically downwards with ever-increasing velocity until it finally reaches the ground; the onward uniform movement parallel to the axis of the piece being continued all the time. We could find the position of the projectile at the end of successive equal periods of time by drawing a straight line AC, a prolongation of the axis of the piece, or a line of the same inclination; on this we mark off equal distances representing by scale the velocity of the projectile per second, the points B, C, D, E being the positions the projectile would be in at the end of each successive second if gravity did not act. In order to bring the diagram within moderate compass, we suppose the projectile to have only the small velocity of 115 ft. per second. At the end of the first second it would be at B, but now suppose that gravity is allowed to act for one second, it would at the end of that time have fallen 16 ft. vertically below B and have arrived at _b_. Similarly we may set off by scale on verticals through C, D, and E distances representing 64 ft., 144 ft., and 256 ft. respectively. Because, for instance, the ball, without gravity acting, would at the end of 3 seconds be at D, where we may suppose its course arrested and gravity then allowed to act for 3 seconds to pull the ball down from its position of rest at D; at the end of this period, gravity alone acting, its position would be 144 ft. vertically below D, because gravity pulls a body that distance in 3 seconds, and the actual position 3 seconds after the ball had left the muzzle would be at _d_, after it had described the curved path A, _b_, _c_, _d_. Supposing _d_ to be the highest point of the trajectory, another 3 seconds would bring the ball along a downward curve, and at the end of 6 seconds from the discharge it would be at a point on the same level as A. Now the complete curve would be symmetrical on each side of a vertical line through its highest point, and it would be in fact a regular _parabola_ with its vertex at _d_.