resistances: then √3.53 1.88, and 175 × 1.88 = 330, being the greatest velocity that can be acquired by the 6 lb ball, and with which it will afterwards uniformly descend. For a 9lb ball, whose diameter is 4:04, the velocity will be 175 × 2.01 = 353. And so on for any other size of iron ball, as in the following table. Where the first column contains the weight of the balls in lbs; the 2d their diameters in inches; the 3d their velocities to which they nearly approach, as a limit, and therefore called their terminal or last velocities, with which they afterward descend uniformly; and the 4th or 18 Term. Wt. Diam. Height inch. Veloc. due to v, feet. feet. 1.94 244 930 2 2.45 275 1182 3 But it is manifest that the balls can never attain exactly to these velocities in any finite time or descent, being only the limits to which they continually approach, without ever really reaching, though they arrive very nearly at them in a short space of time; as will appear by the following calculation. : To obtain general expressions for the space descended, and the time of the descent, in terms of the velocity v: put x = any space descended, t its time, and the velocity acquired, the weight of the ball w = 1.05 lb. Now, by the theorem near the end of prob. 2, which is the proper rule for this case, the velocity being small, '00001725v2 = cv2 is the resistance due to the velocity v; theref. wc is the impelling force, and =f the accelerating force; conseq. vv or fluent of which, by the 8th form, is x = the general value of the space r descended. Here it appears that the denominator w cv decreases increases; conseq. the whole value of x, the descent, increases with v, till it becomes infinite, when the resistance cu2 is the weight of the ball, when the motion becomes uniform, uniform, as before remarked. We may however easily assign the value of r a little before the velocity becomes uniform, or before cu2 becomes w. Thus, when co2=w, then 246, as found in the beginning of this problem. Assume therefore v a little less than that greatest velocity, as for instance 240: then this value of v substituted in the general formula for above deduced, gives x 2781 feet, a little before the motion becomes uniform, or when the velocity has arrived at 240, its maximum being 246. In like manner is the space to be computed that will be due to any other velocity less than the greatest or terminal velocity. On the contrary, to find the velocity due to any proposed spacer, from the formula x = x h. 1. 4gc to = h. l. take therefore the number to the hyp. log. of which w ; conseq. NW - Ncv2 — W, wNcv2, and v = N-1 = -w, a general theorem for the value of v due to any distance. Suppose, for instance, a is 1000. Now 4g being 64, w= 1·05, and c = '00001725; theref. 4gc - 10514, and the natural 20 number belonging to this, considered as an hyp. log. is 2.8617 N; hence then v√. w=199, is the velo N-1 NC city due to the space 1000, or when the ball has descended 1000 feet. Again, for the time t of descent: here t ==; but value of the time t for any value of the velocity v; which value of t evidently increases as the denominator decreases, or as the velocity v increases; and consequently the time is infinite when that denominator vanishes, which is when v or co2w, the resistance equal to the ball's weight, being the same case as when the space a becomes infinite, as above remarked. But, like as was done w. for for the distance x as above, we may here also find the value of t corresponding to any value of v, less than its maximum 246, and consequently to any value of x, as when v is 240 for instance, or x = 2781, as determined above. Now, by substituting 240 for v, in the general formula that it would be nearly 16 seconds when the velocity arrives at 240, or a little less than the maximum or uniform degree, viz, 246, or when the space descended is 2781 feet. Also, to determine the time corresponding to the same, or when the descent is 1000 feet, or the velocity 199: find the feet, or when the velocity is 199. See other speculations on this problem, in the 2d volume, prob. 22, as determined from theory, viz, without using the experimented resistance of the air. PROBLEM X. To determine the Circumstances of the Motion of a Ball projected Horizontally in the Air; abstracted from its Vertical Descent by its Gravitation. Putting d for the diameter, and w the weight of the ball, v the velocity of projection, and the velocity of the ball after having moved through the space r. Then, by corol. 1 to prob. 2, if the velocity is considerable, such as usual in practice, the resistance of the ball, moving with the velocity v, is (mv2 — nv)d2, and therefore mv2- nv force f; hence the common formula W dis the retardive 33d2 mv2. -NO -d, and theref. * = X พ W 32md the fluent of which is obviously × - hyp. log, of i-, and by the correction by the first general formula for the distance passed over in terms of the velocity. Now, for an application, let it be required first, to deter mine in what space a 24lb ball will have its velocity reduced from 1780 feet to 1500, that is, losing 280 feet of its first velocity. Here, d = 5·6, w= 24, v = 1780, and v=1500; 150. Hence =3587-4, then x = 3587·4 × also w m V 150 h. 1. 32md = 3587-4 x h. 1. V 150 676 feet, the space passed over when the ball has lost 280 feet of its motion. Again, to find with what velocity the same ball will move, after having described 1000 feet in its flight. theorem is r or 1000 The above. บ 150 but the number to the hyp. log. is 1.7416 N suppose; Nv-150N 1630, or Nv = 1630 + 150N, and v = N 150786, the velocity when the ball has Next, to find a theor. for the time of describing any space, the fluent of which, by the 9th form, is t = putting y for the first velocity, and 150 for its value, as before. n m Now, to take for an example the same 241b ball, and its projected velocity 1780, as before; let it be required to find in what time this velocity will be reduced to 786. Here then 1780, v = 786, w = 24, d = 5·6, d2 = 31·36, n=·001; hence 786 1780 32nd2 31.36 21353 the hyp. log. of which is 1099; then 31.36 x 18868' •1099 = 2′′628, the time required. For another example, let it be required to find when the velocity will be reduced to 1000, or 780 destroyed. Here v = 1000, and all the other quantities as before. Then 07449; theref. 31.36 x 07449 178, is the time sought. On the other hand, if it be required to find what will be the velocity after the ball has been in motion during any given time, as suppose 2 seconds, we must reverse the calculation 2" being 150 บ 2 23.916 =23.916× 150; theref. 3916093626 is the hyp. log. of འ་ V the number answering to which is 1·08725 = N city at the end of 2 seconds. The foregoing calculations serve only for the higher velocities, such as exceed 200 or 300 feet per second of time. But, for those that are below 300, the rule is simpler, as the resistance is then, by cor. 2 prob. 2; 00000447d2v2= cd2v2, where d denotes the diameter of any ball. Hence then, employing the same notation as before, =f, and - vv= x h. 1. 32cd2 w == the correct Now, for an example, suppose the first velocity to be 300 v, and the last v = 100, for a 24 lb ball. Then w = 24, d = 5°6, d2 = 31·36, c = '00000447; therefore of which is 1.0986; theref. 1·0986 × 5350 = 5878 = x, is the distance. If the first velocity be only 200 V; then V บ =2, the hyp. log. of which is '69315, therefore 69315 × 5350 3708, the distance. And |