and time along pg = mga PQ ช Hence, the whole time through MPQ is PQ ve + - Wł + 2a - x) 9 ✓ 2gx and that through Mpq is PQ 20 (2+ w 20-x) g 29 2a - x which being identical, are equal. 2s 331. Let (s) be the space required ; then the whole time of descent (t) is expressed by ✓ Now since S = 9.To generally the space described in the 4th second is . (49 = 30) = 11/25 and that described in the last second but 4, is 2. {(t – 4)2 – (t – 5)} = g (2t - 9) ( 2 Hence, by the equation, : 2t 9 :: 1 : 3 it = 15 332. Let t be the time required. Then by the expression for the circumstances of motion on inclined planes, gta sin. 6 2 3 S = х & 을 (32 feet) = 24 feet nearly 8 60 X 2 = 4,9740932 8 333. Let Aa, Bb (Fig. 8) 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. For since TP = TA X TB 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 334. Let AB = a, AC = b, and the distance fallen by the latter body before it is overtaken = x; then 6 a + x is the distance fallen by the former in the same time t with the velocity (v) acquired at B. Hence 6 – a + x = tv + to = tv2ga + . 335. If o be the inclination of the plane, the moving force with which P descends is P - Q sin. O P-Q sin. P + Q Now V = got = g. P-Q sin. O x t the velocity of P acquired P+Q in the time t, the inclination of the plane being any whatever. In the problem, 6 = 30°. 2P - Q 2 (P +Q) :: V=g 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 x = tv + 9 2 d'e 28 = ga = (W ga )' = the square of the velocity Ve acquired down since s = .2 29 a 337. Let a be the inclination of the given plane to the horizon, 8 of that sought; then the moving force is P. sin, a W sin. 0 P sin. a - W sin. A I Х sin. O P sin. a – - W sin. O :: P sin. a sin. 0 W sin.” 0 = max. and putting the differential = 0, we get P sin. sin. 0 = 2W giving the inclination required. = min. a 338. Here the moving force is evidently 1 oz. and ;. the 1 33 9 ť 66 3302 290 24 33 and se 340. Let a be their distance at first, x that after the lapse of the time t; then since the spaces described in that time are tu + 9 f', and tv 9.t 2 2 we have (1) Now when the bodies meet x = 0, and . t = .... if for t in equation (1) we put a otor , there results, after reduction a a the distance required. 34). The moving force being H P by the question, we have L 6 342. Here the moving force is 1 oz. and, by the question, we have, therefore, |