250

THEORY AND PRACTICE

TRACT 37.

15. Again, for the time t of descent: here t — —; but

V

X = sr X -as found above, theref. * = — x ->

*5 t/J — Cl> 3 7 2^ "U3 —« CD*

. 20

1 U' -+ ®

the fluent of which is X h. I. —£— , the general

V-®

C

value of the time £ for any value of the velocity v; which

value of t evidently increases as the denominator ^ -- v

decreases, or as the velocity v increases; and consequentlythe time is infinite when that denominator vanishes, whichis when v — \/~, or cv 1 = w, the resistance equal to theball’s weight, being the same case as when the space x be-comes infinite, as above remarked.

16. But, like as was done for the distance x as above, wemay here also find the value of t corresponding to any valueof v, less than its maximum 252, and consequently to anyvalue of x, as when v is 246 for instance, or x = 2927, asdetermined above. Now, by substituting 246 for v, in thegeneral formula

, w

, I ,«. ui V ~ +V

t== T g ^T x hJ -—-

, it brings out t ~ 16"-957; so

that it would be nearly ]7 seconds when the velocity arrivesat 246, or a little less than the maximum or uniform degree,viz, 252, or when the space descended is 2927 feet.

77. Also, to determine the time corresponding to thesame, w'hen the descent is J000 feet, or the velocity 201:

1-125

find the value of - ^ . ioool76

252 _63

“fir lhen

64

V-

w

a/-7~ v

252 + 201 453

252 - 201 '

51

; the hyp. log. 0 f which is 2-18407.

63

Hence 2-18407 X 16 = 8 " ,6 > the time of descending 1000feet, or when the velocity is 201.

See other speculations on this problem, in the 2d volumeof the Course, prob, 22, as determined from theory, viz,without using the experimented resistance of the air.