theref. 1.50958 × 4256 = 6424r the height, being a little more than a mile. We may now examine what will be the height ascended, considering the resistance always as the square of the velocity. PROBLEM V. To determine the Height ascended by a Ball projected as in the two foregoing problems; supposing the Resistance of the Air to be as the Square of the Velocity. Here it will be proper to commence with selecting some experimented resistance corresponding to a medium kind of velocity, between the first or greatest velocity and nothing, from which to compute the other general resistances, by considering them as the squares of the velocities. It is proper to assume a near medium velocity and its resistance, because, if we assume or commence with the greatest, or the velocity of projection, and compute from it downwards, the resistances will be everywhere too great, and the altitude ascended much less than just; and, on the other hand, if we assume or commence with a small resistance, and compute from it all the others upwards, they will be much too little, and the computed altitude far too great. But, commencing with a medium degree, as for instance that which has a resistance about the half of the first or greatest resistance, or rather a little more, and computing from that, then all those computed resistances above that, will be rather too little, but all those below it too great; by which it will happen, that the defect of the one side will be compensated by the excess on the other, and the final conclusion must be near the truth. Thus then, if we wish to determine, in this way, the alti tude ascended by the ball employed in the 1st table of resistances, when projected with 2000 feet velocity; we perceive by the table, that to the velocity 2000 corresponds the resistance 98 lb; the half of this is 49, to which resistance corresponds the velocity 1400 in the table, and the next greater velocity 1500, with its resistance 574, which will be properest to be employed here. Hence then, for any other velocity v, in general, it will be, according to the law of the 57v2 squares of the velocities, as 15002 2:: 57: 1500% *0000254v2 av2, putting a = '000025, which will denote the air's resistance for any velocity v, very nearly, counting from 2000. Now let a denote the altitude ascended when the velocity is v, and w the weight of the ball: then, as above, av2 is the resistance resistance from the air, hence av + w is the whole resisting w a the fluent of which, by form 8, ish. log. (v2 + =~~); which when x = 0, and v = v the first or projectile velocity, becomes 0= x h. 1. (v2+); theref. by subtract 4ga w a ing, the correct fluent is x = x h. l. 4ga a when the velocity is reduced to v; and when v = 0, or the w 4ga av2+ w x h. 1. velocity is quite exhausted, this becomes Ex. 1. The values of the letters being w=1051b, 4g=64, a=0000254, the last expression becomes 645 x hyp. log. v2+41266 vel. v being 2000, the same expression 1484 x log. 41266 becomes 1484 x log. of 97.93 = 2955 for the height ascended, on this hypothesis; which was 2922 by the former problem, being nearly the same. W Ex. 2. Supposing the same ball to be projected with the velocity of only 1500 feet. Then taking 1100 velocity, whose tabular resistance is 27.6, being next above the half of that for 1500. Hence, as 1100: 2 :: 27·6: 00002375v2=av2. This value of a substituted in the theorem also 1500 for v, and 105 for w, it brings out = 2728 for the height in this case, being but a little above the ratio of the square roots of the velocities 2000 and 1500, as that ratio would give only 2560. 4ga w Ex. 3. To find the height ascended by the first ball, projected with 860 feet velocity. Here taking 600, whose resistance 6.69 is a near medium; then as 6002: 6.69 :: 1 : 0000186a. Hence 2334 the height; which is less than half the range (5100) at 45° elevation, but more than half the range (4100) at 15° elevation, in pa. 161 vol. 2; being indeed nearly a medium between the two. Er. 4. With the same ball, and 1640 velocity. Assume 1200, whose resistance 34 13 is nearly a medium. Then as 12002: 34.13 :: 1 :·0000237=a. Hence พ 64a x h. 1. av2+w =2854; again less than half the range (6000) by experiment in vol. 2, even with 15° elevation. Ex. 5. For any other ball, whose diameter is d, and its weight w, the resistance of the air being = d2v2 3.86 150000 =bd22 bd2x2 + w w -W and x = and bd2v2 + w 28 for the whole height when v = 0. Now if the ball be a 24 pounder, whose diameter is 5'6, and its square 31-36; then being more than double the height of that of the small ball, or a little more than a mile, and very nearly the same as in the 2d example to prob. 4. PROBLEM VI. To determine the Time of the Ball's ascending to the Height determined in the last prob. by the same Projectile Velocity as there given. the time in general when the first velocity v is reduced to V. And when v = 0, or the velocity ceases, this becomes Now, as in the last prob. v=2000, w=1·05, a='000023+ 000000 10 98445 the tangent, to which corresponds the arc 1 of 89° 25', whose length is 1-5606; then × 203·14 x 1:3606= 203·14 × 1·5606 2g = 9"-91, the whole time of ascent. Remark. The time of freely ascending to the same height 2955 feet, that is, without the air's resistance, would be 2953 162955 = 13"-59; and the time of freely ascending, commencing with the same velocity 2000, would be =- =-=-=-= 62°4 = 1′2′′ }. 32 PROBLEM VII. To determine the same as in prob. v, taking into the account the Decrease of Density in the Air as the Ball ascends in the Atmosphere. In the preceding problems, relating to the height and time of balls ascending in the atmosphere, the decrease of density in the upper parts of it has been neglected, the whole height ascended by the ball being supposed in air of the same density as at the earth's surface. But it is well known that the atmosphere must and does decrease in density upwards, in a very rapid degree; so much so indeed, as to decrease in geometrical progression, at altitudes which rise only in arithmetical progression; by which it happens, that the altitudes ascended are proportional only to the logarithms of the decrease of density there. Hence it results, that the balls must be always less and less resisted in their ascent, with the same velocity, and that they must consequently rise to greater heights before they stop. It is now therefore to be considered what may be the difference resulting from this cir cumstance. 1 Now, the nature and measure of this decreasing density, of ascents in the atmosphere, has been explained and determined in prop. 76, pa. 244, &c, vol. 2. It is there shown, that if D denote the air's density at the earth's surface, and dits density at any altitude a, or r; then is x = 63551 × log. of in feet, when the temperature of the air is 55°; for the temperature of freezing cold; we ratio of ea to cz: thereon resistance at the height, te veloc 0000254. To this adding, the wegn & th c-a c+I tw for the whole resistance pot fret to and the ball's mass; conser. wil demote the accelerating force of the ball. Or, if we include the w small part * or 1, within the factor which will make พ [ no sensible difference in the result, but be a great deal simpler in the process, then is + x =ƒ the accelerating |