similar and equal to the portion BE of the parabola BEP, but turned the contrary way. Conceiving either the two curves AE and EP, or the continued curve BEP, to be described by a projectile in its motion, it is manifest that, whether the greater portion of the curve be described before or after the ball reaches the wall DR, will depend on its initial velocity, and on the distance AC or BC, and on the angle of projection. The problem then is now reduced to this, viz, To find the angle at which a ball shall be projected from B, with a given impetus, so that the distance DP, at which it falls, from the given point D, on the plane DP, parallel to the horizon, shall be a maximum. Now this problem may be constructed in the following manner: From any point E in the horizontal line DC, let fall the indefinite perp. EG, on which set off EB = the impetus corresponding DE B to the given velocity, and BI = 24 the distance of the horizontal plane below the point of projection; also, through i draw AP parallel to DC, From the point B set off BP = BE + EI, and bisect the angle EBP by the line Bн: then will BH be the required direction of the ball, and IP the maximum range on the plane AP. For, since the ball moves from the point B, with the velocity acquired by falling through EB, it is manifest, from p. 156 vol. 2, that DC is the directrix of the parabola described by the ball. And since both B and P are points in the curve, each of them must, from the nature of the parabola, be as far from the forces as it is from the directrix; therefore B and P will be the greatest distance from each other when the focus F is directly between them, that is, when BP = BE + CP. And when BP is a maximum, since BI is constant, it is obvious that IP is a maximum too. Also, the angle FBH being EBH, the line BH is a tangent to the parabola at the point B, and consequently it is the direction necessary to give the range IP. Cor. 1. When B coincides with I, IP will be = BP = BE + EI = 2EI, and the angle EBH will be 45°: as is also manifest from the common modes of investigation. Cor. 2. When the impetus corresponding to the initial velocity of the ball is very great compared with AC or BC (fig. 1), then the part AE of the curve will very nearly coincide with its tangent, and the direction and velocity at a may be accounted the same as those at E without any sensible VOL. III. Y error. error. In this case too the impetus BE (fig. 2) will be very great compared with BI, and consequently, B and I nearly coinciding, the angle EBH will differ but little from 45°. Calcul. From the foregoing construction the calculation will be very easy. Thus, the first velocity being 80 feetv, 99-48186 BE the imthen (vol. 2 pa. 156) 4g 80 x 80 petus; hence EI= FP = 101.98186, and BP = BE + EI = 201-46372. Now, in the right-angled triangle BIP, the sides BI and BP are known, hence IP 2014482, and the angle IBP 89° 17′ 20′′: half the suppl. of this angle is 45° 21′ 20′′. 10= = EBH. And, in fig. 1, IP — ID = 201·4482 191.4482 = DP, the distance the ball falls from the wall after reflection. PROBLEM 17. From what height above the given point a must an elastic ball be suffered to descend freely by gravity, so that, after striking the hard plane at B, it may be reflected back again to the point A, in the least time possible from the instant of dropping it? Let c be the point required; and put AC, and AB = a; then \√ CB = √(a + x) is the time in CB, and CA is the time in CA; therefore ‡√(a + x) − ž√x is the time down AB, or the time of rising from B to A again: hence the whole time of falling through CB and returning to A, is √(a+x) √x, which must be a min. or 2√(a + x)−√ x = 0, and hence a minimum, in fluxions(+1) 27x *=ja, that is, ac = AB. PROBLEM 18. B Given the height of an inclined plane; required its length, so that a given power acting on a given weight, in a direction parallel to the plane, may draw it up in the least time possible. Let a denote the height of the plane, x its length, p the power, and the weight. Now the tendency down the plane aw aw (p+w) x the accelerating force ƒ; hence, by the theorems for constant forces, pa. 342 vol. 2, t = (p+w)x2 must be be a minimum, or pr - aw a min.; in fluxions, 2(px-aw)xx pr2x = 0, or pr=2aw, and hence p: w:: 2a :x :: double the height of the plane to its length. PROBLEM 19. A cylinder of oak is depressed in water till its top is just level with the surface, and then is suffered to ascend; it is required to determine the greatest altitude to which it will rise, and the time of its ascent. Let a the length, and b the area or base of the cylinder, m the specific gravity of oak, that of water being 1, also any variable height through which the cylinder has ascended. Then, a being the part still immersed in the water, (ax) xbx1 = (a-x)b is the force of the water upwards to raise the cylinder; and a x b x m = abm is the weight of the cylinder opposing its ascent; therefore the efficacious force to raise the cylinder is (a-x)b — abm; and, the mass being abm, the accelerating force is (a-x)b- abm a-x- am abm = am putting n = 1-m the difference between the specific gravities of water and oak. Now if v denote the velocity of ascent at the same time when a space is ascended, then by the theorems for variable forces, vv= 32fx 82 am 32 = x (anxxx), therefore am 2am 22 = × (2anx x2), and v = 8 √· : but when : 0, the cylinder has acquired its greatest ascent, v and v2 = therefore 2anx r2=0, and hence x = 2an the part of the cylinder that rises out of the water, being 15a or its length. of To find when the velocity is the greatest, the factor 2anx in the velocity must be a max. then 2anx 2xx = 0, and xan, being the height above the water when the`velocity is the greatest, and which it appears is just equal to the half of 2an above found for the greatest rise, when the upward motion ceases, and the cylinder descends again to the same depth as at first, after which it again returns ascending as before; and so on, continually playing up and down to the same highest and lowest points, like the vibrations of a pendulum, the motion ceasing in both cases in a similar manner at the extreme points, then returning, it gradually accelerates till arriving at the middle point, where it is the greatest, then gradually retarding all the way to the next extremity of the vibration, thus making all the vibrations in equal times, to the same extent between the highest and lowest points, except that, by the small tenacity and friction. &c. of the water against the sides of the cylinder, it will be gradually and slowly retarded in its motion, and the extent of the vibrations decrease till at length the cylinder, like the pendulum, come to rest in the middle point of its vibrations, where it naturally floats in its quiescent state, with the part na of its length above the water. The quantity of the greatest velocity will be found, by substituting na for r, in the general value of the velocity , when it becomes 8n a very nearly, Lanx *2 2am a 2m 8√ the value of m being '925, and consequently that of n = 1− m = '075. To find the time t answering to any space x. Here circular arc to radius 1 and versed sine Now at the mid na na, and then the vers. I na na na dle of a vibration r is the radius, and A is the quadrantal arc 1.5708; then the flu. becomes√2ma × 1∙5708 =∙17√/a × 1·5708= •267/a for the time of a semivibration; hence the time of each whole vibration is 534/a = √a, which time therefore depends on the length of the cylinder a. To make this time = 1 second, a must be () very nearly 34 feet or 42 inches. That is, the oaken cylinder of 42 inches length makes its vertical vibrations each in 1 second of time, or is isochronous with a common pendulum of 394 inches long, the extent of each vibration of the former being 6 inches. = PROBLEM 20. = Required to determine the quantity of matter in a sphere, the density varying as the nth power of the distance from the centre? Let r denote the radius of the sphere, d the density at its surface, a = 3·1416 the area of a circle whose radius is 1, and any distance from the centre. Then 4ar will be the surface of a sphere whose radius is a, which may be considered by expansion as generating the magnitude of the solid; therefore 4axx will be the fluxion of the magnitude; but as tity of the matter in the whole sphere. Corol. 1. The magnitude of a sphere whose radius is r, being fur3, which call m; then the mass or solid content will be xm, and the mean density is 3d n+3 3d Corol. 2. It having been computed, from actual experiments, that the medium density of the whole mass of the earth is about 5 times the density d at the surface, we can now determine what is the exponent of the decreasing ratio of the density from the centre to the circumference, supposing it to decrease by a regular law, viz, as "; for then it will be 5d = and hence n = — 2. So that, in this 3d n+3 r 2 1 case the law of decrease is as 3, or as that is, inversely as the ths power of the radius. PROBLEM 21. Required to determine where a body, moving down the convex side of a cycloid, will fly off and quit the curse. Let AVEB represent the cycloid, the properties of which may be seen at arts. 146 and 147 vol. 2, and VDC its generating semicircle. Let E be the point where the motion com H C BG mences, whence it moves along the curve, its velocity increasing both on the curve, and also in the horizontal direction DF, till it come to such a point, F suppose, that the velocity in the latter direction is become a constant quantity, then that will be the point where it will quit the cycloid, and afterwards describe a parabola FG, because the horizontal velocity in the latter curve is always the same constant quantity, by art. 76 vol. 2. Put the diameter vcd, vнa, vi=x; then VD=√dx, and ID= ✔(dx-x2). Now the velocity in the curve at F in descending down EF, being the same as by falling through HI or л-a, by art. 139, will be = 8(x-a); but this ve locity |