Now, to find how near to the Velocs. Comput. Exper. truth this theorem comes, in every instance in the table, by or v. resists resists 700 9.90 9.44 800 13.38 12.81 900 17.37 16.94 1000 21.88 21.88 1100 26.90 27:63 1200 32:44 34.13 a ball, in all degrees of its velocity, from the first or greatest, ··00388v=r, Corol. 1. The foregoing rule ·00002576v2 =(0000066703-001v)da (000002001v)d2, which Corol. 2. And in a similar manner, to adapt the theorem dd of ball, we must multiply it by the ratio of the surfaces, 3.86 by which it becomes 00000447d2v2=r. We shall soon take occasion to make some applications in the use of the foregoing formulas, after considering the effects of such velocities in the cases of nonresistances. PROBLEM III. To determine the Height to which a Ball will rise, when fired from a cannon Perpendicularly Upwards with a Given Velocity, in a Nonresisting Medium, or supposing no Resistance in the Air. By art. 73, Motion and Forces, this vol. it appears that any body projected upwards, with a given velocity, will ascend to the height due to the velocity, or the height from which it must naturally fall to acquire that velocity; and the spaces fallen being as the square of the velocities; also 16 feet being the space due to the velocity 32; therefore the space due to any proposed velocity v, will be found thus, as 322 : 16 :: v2: s the space or as 64: 1 :: 2: 2s the space, or the v22s 4 height to which the velocity v will cause the body to rise independent of the air's resistance. 6.4 64 Exam. For example, if the first or projectile velocity, be 2000 feet per second, being nearly the greatest experimented velocity then the rule s becomes 20003 625,00 feet 115 miles: that is, any body, projected with the velocity 2000 feet, would ascend nearly 12 miles in height, without resistance. Corol. Because, by art. 88 Projectiles this vol. the greatest range is just double the height due to the projectile velocity, therefore the range at an elevation of 45°, with the velocity in the last example, would be 233 miles in a nonresisting medium. We shall now see what the effects will be with the resistance of the air. ร PROBLEM IV. To determine the Height to which a Ball projected Upwards, as Putting to denote any variable and increasing height as-" cended by the ball; v its variable and decreasing velocity there; d the diameter of the ball, its weight being w; m = '00000, and n = 001, the co-efficients of the two terms denoting the law of the air's resistance. Then (mv2 - nv)d2, by cor. 1 to prob. 17 prob. 2, will be the resistance of the air against the ball in avoirdupois pounds: to which if the weight of the ball be added, then (mv2 — nv)d2+w will be the whole resistance to the ball's motion; this divided by w, the weight of the ball in (mv2 — nv)d2 + w motion, gives W W d2+1=f the re tarding force. Hence the general formula v=2gfx (theor. 10 pa. 379 this volume) becomes →vp=2gxx (m22 (mv2 — nv)d2 +21 making negative because v is decreasing, where g 16 ft.; Now, for the easier finding the fluent of this, assume v — 22 in2 n =z2, and v2 m 4m2 ; these being substituted in the above value of x, it be putting p= 2m2 W + and q2= -p2, or p2 + q3= md2 Then the general fluents, taken by the 8th and 11th forms of the table of Fluents give x= + X arc to rad. q and tang. v-p]. But, at the beginning qa of the motion, when the first velocity is v for instance, and the space x is = 0, this fluent becomes v-p.] Hence by subtraction, and taking v = 0 for the end of the motion, the correct fluent becomes x= (v3 — v+ )-log. +x (arc tan. v-p- arc tan. -p to rad. q)]. md2 q3 P But as part of this fluent, denoted by X the dif. of the two arcs to tans. v p and q2 most height to which the ball will ascend, when its motion But now, for the numerical value of the general coefficient 4gmd and the term ; md2 because the mass of the ball to the diameter d, is 5236sd3, if its specific gravity be s its weight will be ⚫5236sd3w; therefore 78540sd, this divided by 4g or 64 it gives mda 1227.2sd for the value of the general coefficient, to any diameter d and or 760d X com. log. Exam. 1. Suppose the ball be that belonging to the first i v, the height ascended is found to be 2920 feet, or little more than half a mile; though found to be almost 12 miles without the air's resistance. And thus the height may be found for any other diameter and velocity. Exam. 2. Again, for the 24lb ball, with the same velocity 2000, its diameter being 56 d. Here 760d = 4256; y2 150v+21090d 38181 the log. of which is 1.50958; and 21090d 1181' theref. 1.50958 X 4256 = 6424=x 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 every where 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 altitude ascended by the ball employed in the 1st table of resistauces when projected with 2000 feet velocity; we perceive by the table, that to the velocity 2000 corresponds the resistance 981b; 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 571, which will be properest to be employed here. Hence then, for any other velocity |