Having obtained the time, it will be easy to work out the lengths _b_ and _a_ as 26,648 ft. and 9114·1 ft. respectively; and as _c_/_b_ = _cosine_ 20°, we have _c_ = 26,648 × ·9396926 = 25040·8 ft., which is the _range_. The trajectory will be a curve (parabola) symmetrical on each side of a vertical line half-way between A and B, and the length of this line within the triangle will be equal to half of _a_, and in half of 23·7927 seconds the projectile, supposed to move only along the line AC, would reach the point where this vertical axis intersects AC. If during this half-time it had been falling from rest at the same intersection, it would have reached a point below by a space just one quarter of CB (the spaces fallen through being as the _squares_ of the times), and therefore at this its highest point its distance above AB would also be one quarter the length of _a_ = 2278·525 ft., which distance is called the _height_ of the trajectory; and the descending curve being in every respect symmetrical to the ascending branch, the angle at which this would be inclined to AB would be 20°, but in the opposite direction to BAC, while the velocity would be the same as at A. We may now compare these results with those calculated when the air resistance is taken into account:— ┌─────────────────────────────────┬───────────┬───────────┬───────────┐ │ │Without air│With air │ │ │ │resistance.│resistance.│Difference.│ ├─────────────────────────────────┼───────────┼───────────┼───────────┤ │Time of descent │23·7927 │22·61 sec. │–1·18 sec. │ │ │secs. │ │ │ │Angle of descent │20° │23° 49´ │+3° 49´ │ │Velocity of descent │1120 │868·8 │–251·2 │ │ │foot-secs. │foot-secs. │f.-s. │ │Range │25040·8 ft.│20,622 ft. │–4418·8 ft.│ │Height of trajectory │2278·5 ft. │1989 ft. │–288·5 ft. │ └─────────────────────────────────┴───────────┴───────────┴───────────┘ With the air resistance the trajectory will no longer be a symmetrical curve: its highest point, instead of being on the vertical line midway between A and B, will be on one 1,050 ft. nearer to B than to A, and the descending branch will be steeper than the ascending. The total time, it will be observed, is less, although the final, and therefore the mean, velocity, is also less; but this shortening of the time is due to the trajectory itself being much less in length. The range of the projectile is decreased by 4,418 ft., or 1,473 yards, or more than four-fifths of a mile. The loss of velocity at the descent is very notable, and the reader will find it interesting to calculate the corresponding loss of energy by the formula already given.