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Now the fluent of the first side of this equation is evidently 2c × h. l. (c+x); and the fluent of the latter

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side, the same as in prob. 5, is × h. I. (v2 + ); there





fore the general fluential equa. is-x+2c x h. 1. (c+x)= × h. 1. (2). But, when a = 0, and v=v the initial velocity, this becomes 0+2cx h. 1. (v2+); theref. by subtraction, the correct fluents are x + 2c x h.l. when the first velocity v is diminished to any less one v; and when it is quite extinct, the




x h. 1.



av2 + w

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h. l. ave+, for the greatest height x ascended.

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Here, in the quantity h. 1. +, the term x is always small


in respect of the other term c; therefore, by the nature of

logarithms, the h.l. of is nearly

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Now the latter side of this equation is

the same value for r as was found in the 5th problem, which therefore put = b; then the value of x will be easily found from the formula xb, by a quadratic equation. Or, still easier, and sufficiently near the truth, by substituting b for r in the numerator and the denominator of then


2c -x


20-b b, and hence x = 2c+bb, or by proportion, as

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2cb: 2c+b::b: x; that is, only increase the value of r, found by prob. 5, in the ratio of 20 b to 2c + b.

Now, in the first example to that prob. the value of x or b was there found 2955; and 2c being = 108000, theref. 2c - b⇒ 105045, and 2c + b = 110955, then as 105045 : 110955: 2955: 3121 = the value of the height x in this case, being only 166 feet, orth part more than before.


Also, for the 2d example to the 5th prob. where x was = 6420; therefore as 2c-b : 2c + b or as 105045 :'110955 :: 6420: 6780 the height ascended in this example, being also the 18th part more than before. And so on, for any other examples; the value of 2c being the constant number 108000.


To determine the Time of a Ball's Ascending, considering the Decreasing Density of the Air as in the last prob.

The fluxion of the time is t =. But the general êquation of the fluxions of the space r and velocity v, in the last

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ways small in respect of c, is nearly =b as determined in the

· + ს c-b


last problem; theref. may be substituted for with

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Now, this fluxion being to that in prob. 6, in the constant ratio of cb to c + b, their fluents will be also in the same constant ratio. But, by the last prob. c=54000, and b=2955 for the first example in prob. 5; therefore c-b51045, and cb56955, also, the time in problem 6 was 9"-91; therefore as 51045: 56955 :: 9′′·91 : 11′′·04 for the time in this case, being 1"13 more than the former, or nearly the 9th part more; which is nearly the double, or as the square of the difference, in the last prob. in the height ascended.


To determine the circumstances of Space, Time, and Velocity, of a Ball Descending through the Atmosphere by its own Weight.

It is here meant that the balls are at least as heavy as cast iron, and therefore their loss of weight in the air insensible; and that their motion commences by their own gravity from a state of rest. The first object of enquiry may be, the utmost degree of velocity any such ball acquires by thus descending. Now it is manifest that the ball's motion is commenced, and uniformly increased, by its own weight, which is its constant urging force, being always the same, and producing an equal increase of velocity in equal times, excepting for the diminution of motion by the air's resistance. It is also evident that VOL. III. this


this resistance, beginning from nothing, continually increases, in some ratio, with the increasing velocity of the ball. Now, as the urging force is constantly the same, and the resisting force always increasing, it must happen that the latter will at length become equal to the former: when this happens, there can afterwards be no further acceleration of the motion, the impelling force and the resistance being equal, and the ball must ever after descend with a uniform motion. It follows therefore that, to answer the first enquiry, we have only to determine when or what velocity of the ball will cause a resistance just equal to its own weight.

Now, by inspecting the tables of resistances preceding prob. 1, particularly the 1st of the three tables, the weight of the ball being 105 lb, we perceive that the resistance increases in the 2d column, till 069 opposite to 200 velocity, and 1.56 answering to 300 velocity, between which two the proposed resistance 105, and the correspondent velocity, fall. But, in two velocities not greatly different, the resistances are very nearly proportional to the squares of the velocities. Therefore, having given the velocity 200 answering to the resistance 0·69, to find the velocity answering to the resistance 1.05, we must say, as 0.69: 1:05 :: 2002: v2 = 60870, theref. v=60870 246, is the greatest velocity this ball can acquire; after which it will descend with that velocity uniformly, or at least with a velocity nearly approaching to 246.

The same greatest or uniform velocity will also be directly found from the rule 00001725v2r, near the end of problem 2, where r' is the resistance to the velocity v, by making 1.05= r; for then v2 = for vas before.

1.05 .00001725

-=60870, the same value

But now, for any other weight of ball; as the weights of the balls increase as the cubes of their diameters, and their resistances, being as the surfaces, increase only as the squares of the same, which is one power less; and the resistances being also in this case, as the squares of the velocities, we must therefore increase the squares of the velocity in the ratio of the diameters of the balls; that is, as 1.965:

d: 2462: dvand hence v = 246 √ = 175/d.

2462 1.965

d 1.965

If we take here the 3 lb ball, belonging to the 2d table of resistances, whose diameter d is 2.80; then √✓✅/2·80=1·673, and 175 x 1·67 = 294, is the greatest or uniform velocity, with which the 3 lb ball will descend. And if we take the 6lb ball, whose diameter is 3.53 inches, as in the 3d table of


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







Veloc. due to v,







2.45 275



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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

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fluent of which, by the 8th form, is x =

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the general value of the space r descended.

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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,

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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.

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= h. l.

take therefore the number to the hyp. log. of which

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; conseq. NW - Ncv2 — W,

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wNcv2, and v =



-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


number belonging to this, considered as an hyp. log. is 2.8617 N; hence then v√. w=199, is the velo



city due to the space 1000, or when the ball has descended 1000 feet.

Again, for the time t of descent: here t ==; but

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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



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