Why? The Science of Athletics

260 WHY?-THE SCIENCE OF ATHLETICS Then (i) v =gt; or t =;;secs. (ii) Now h =igt 2 • 252 252 .. h =i X 32 X - 2 = 6 -ft. 32 4 _ =9·76 ft. or 9ft. 9 ins. But this height obtained applies to the centre of gravity of the body in question. Assuming the centre of gravity of a man when running is at the mid-point of a line drawn between the upper ends of his femur it would be approximately 3 ft. from the ground and the line of his take-off, and should be raised by merit of his speed of 25 ft. per second to a height of 9 ft. 9 ins. +3 ft. = 12 ft. 9 ins. If, however, the athlete is capable of running roo yards in r r secs., i.e. his approach run is made at a speed of 27·3 ft. per sec., then the height his found to be r I ·7 ft., so that he should be able to raise his centre of gravity- to and propel his body over a cross-bar raised 14ft. 7 ins. ·above the ground. The foregoing figures relate purely and simply to the merit of the speed of the run-up, and such factors as friction of the pole-point in the slide-way and air-resistance have not been allowed for, although the latter is a very large factor indeed, which must have an adverse effect upon the height obtained. On the other hand, no favour– able allowance has been made for the vertical lift (shown in progress in Fig. 45, Plate II) whereby the vaulter raises his body a considerable way above the height of his hand-hold upon the pole when it has reached a vertical position. Against that, as we shall see presently, a man accomplishing a high jump of 6 ft. has to raise his centre ofgravity approximately to 6 ft. 5 ins. to get a safety margin in clearance. None the less, in the Pole Vault, the con– version of a strong and rapid pull into the action of a quick and forceful push-up should be worth at least r8 ins., especially if maximum foot-elevation has been obtained.