:: x = 2ň ab .. the time of B's uniform motion is 2N ab 2α the time of A's descent. 2gb 9 v 403. Let o be the inclination of the line to the horizon, then the time down it is - 28.02 v2 = x3 But by the question, X = 300 x cos. 30° = 150 3 and y – 300 X sin. 30o = 150 whence v = 15 2g. Hence the greatest altitude above the plane is easily found, by the rules of max. and min., to be 05 feet; and the time of flight 150 N 3 = 10 V cos. 60 405. Let a be the inclination of the given plane to the horizon, a the vertical distance between the given point and the plane; also suppose the co-ordinates (x, y) of the required locus to originate in the intersection of this vertical with the plane, s * being measured I and y parallel to the intersection of the plane with the horizon; then (t', y) being the horizontal and vertical co-ordinates of the path, described by the body projected parallel to the horizon, we easily obtain y' = a - 2 sin. a 2'2 = y + xocos. « And by 370, gx 20% C Hence 20cos. Sa g the equation to an ellipse, whose transverse and conjugate axes are 2v 2v 2ag + va tan.'a, and 2ag +vo tan.'« respectively. g cos. a g the equation to a circle, whose radius is tv, and the co-ordinates of whose centre are O and tv gt? Let p be the parameter of the given parabola, whose 408. equation is = py 1 *, and y being the horizontal and vertical co-ordinates, originating in the vertex; then, supposing (a,b), (a', ') the co-ordinates of the given points of projection, the 2 (a, a) of projection, or the inclinations of the tangents to the parabola, at those points are dy = tan. O) determined by And :: v' = 2gb (1 + eos. ) = 2gb (1 + d'? = 2gb (1 + cos, 'a') = 2gb' (1 + 4a'! Also, at the point of concourse, a - x = tv cos. a, a + x = tv' cos. cé = 409. Let m denote the degree of elasticity; then, 0, 0, X, being the 2 of reflection made with the radii, we have, (Wood, p. 131.) tan. 0 = m tan. tan.Q = m tan. x 2 410. Let o be the inclination of the plane AB to the horizon, a, b the horizontal and vertical co-ordinates of the point A referred to the point B; then, v denoting the velocity at B, or that with which the body is reflected by the vertical plane in a direction inclined to the horizon by the 20, the trajectory described, referred by (x, y) to B, will be (370). g.ro y = x tan. A 2v2 cos. 20 But, by the question, the body strikes the plane again in A;:. - b = a tan. 0 ga? But b = a tan. 0, a = N a? +63 x cos. 0. ag 2 sin.20 or 411. Let V, V, V .....denote the velocities of projection after each rebound, 0, 04, 0.......the elevations of projection. t, ti, to .......the times of flight. 7, 71, 7 ....... the horizontal ranges. h, hl, he......the greatest altitudes, a, Q ,, ....... the areas of the parabolas. VOL. II. T cos.bg Then, since the velocity of incidence on the horizontal plane is = that of the reflection immediately preceding, and (Wood, p. 131.) cos. O cos. , V, Ug = &c. = &c. cos. and tan. 0, = n tan. 0, tan. 0, = n tan. 01, &c. = &c. :: (370) sin. 20, r, = 0, sin. 20, &c., become by substitu2g 29 tion, and reduction. |