344. Let x be the weight required. Then since P = Q, the moving force = x, and p = 2P + 2 Hence v = got, will give, by the question, 48 = g x 2P+x X 6. 16P and .. I 9-8 the weight required. 9 Qto = = 144 feet. 345. Let x denote the required weight. Then the moving force is P and p = P + x P-X Let F, P be the accelerating forces, then by the 346. question, 25 36 F 5 Hence, if Q and Q' be the quantities of matter, the ratio of the moving forces is 5 F x Q 125 FXQ 144 348. Let m be the degree of elasticity, and the altitude required; also let a be the given velocity of projection. Then the velocity with which the body reaches the plane is (a + 2gx), and that with which it quits the plane is mn(a + 29x). VOL. II. R But by the question z is the space due to this velocity, and the time is given, :: mo (a + 2gx) 29 2g 349. Let a be the radius of the circle, (at', y'); (f, y) the corresponding co-ordinates of the circle and locus measured from the horizon along the vertical diameter; then in any time t, considering the chord v r'? + y's an inclined plane whose height is t', the form 9 H ť X I X gives w x + y' N 2 + y = gcu W x? +y also y S = 2 2 19 and y'a = 2ax x'? whence eliminating a', y', &c., we have 4a - gt ya x2 2 4a – gť or the locus is a circle whose radius is 4 350. Let a, b, be the altitudes fallen by A, B, and x the space described along the horizon (which they will describe uniformly with the velocities acquired, since the plane of reflection pro jects then along the horizon). Then, the velocities acquired being ✓ 2ga, and 2gb if the whole time of motion be denoted by t, we have 2 b + 9 ✓ 2gb :: x = 2 Nab. Now, if a circle pass through the points of departure and touch the horizontal line, the distance, as is well known, of the point of contact from the concourse of the tangent and line passing through those points = 7ab. .. if a circle, &c. &c. Q. E. D. 351. Let t, t, be the times down the chords corresponding to ordinates y, y', and abscissæ x, r'. Then considering the chords, planes inclined to the horizon by the < 45° and 15°, we 9 have, by the form s = H L H 2 But x = ข 12+ 3 tan. 15° 2-13 v. (2 + N 3) (2+73)" + 1 = 4y W x.(20 – x) : 2 '20 — a :: x.(2-x): 2 x'.(2-x) a being the semi-axis. •352. force is Let x be the power required. Then the moving |