331. = 2 +√ Qa−x) Let (s) be the space required; then the whole time of descent (t) is expressed by 2. (2t - 9) and that described in the last second but 4, is L. {(t — g 2 { (t − 4)2 — (t − 5)o} = g 332. Let t be the time required. Then by the expression for the circumstances of motion on inclined planes, 333. Let Aa, Bb (Fig. 88) be the perpendiculars to the horizon bT. Then, joining AB let it be produced to meet ba in T, and take TP a mean proportional between TA, TB, and describe the circle APB passing through A, P, B, and join AP, PB, P is the point required. the circle touches the horizon in P. Hence time down BP time AP. The point B will fall within or without ab according as the arc or < 10. PAB is > or 334. Let AB = a, AC = 6, and the distance fallen by the latter body before it is overtaken = x; then b-a + x is the distance fallen by the former in the same time t with the velocity (v) acquired at B. Hence 335. If be the inclination of the plane, the moving force in the time t, the inclination of the plane being any whatever. 336. Let t be the time in which the bodies meet, a the distance between the points of projection, and from the upper of these let x be the distance of the point of concourse; then THE GRAVITATION OF BODIES. = v'2 — v2 = ga = (ga) the square of the velocity 337. Let a be the inclination of the given plane to the horizon, of that sought; then the moving force is P. sin. α- W sin. and the accelerating force (9) is P P sin. a W sin. 0 P+W 338. Here the moving force is evidently 1 oz, and.. the accelerating force is @ = by avoirdupoise weight. 1 33 |