TRACT 37.

OP GUNNERY.

233

81‘. For another example, let it be required to find whenthe. velocity will he reduced to 1000, or 780 destroyed. Herev = 1000, and all the other quantities as before. Thenv-231 ^ _ v _

1369

231

X — =

1519 1000 1549 ^ , r i • . •

x , the log. or which is

769

1780 '

•053648; theref, 32f X -053648 = T723 seconds, is the timesought.

8.5. On the other hand, if it be required to find what willbe the velocity after the ball has been in motion during anygiven time, as suppose 2 seconds, we must reverse the cal-culation thus: t — 2" being = S2x X log.

theref.

•062218 is the log. of

v - 231

v -131 _

v — 231 ' v ’the number

39X — .“t>‘ ' v — 231 ’ »•’

answering to which is 1 - 1536 = N suppose, that is, n =

v — 231 V

Hence nv» — 231nv — yv — 23 Lr, and v == 941, the velocity at the end of 2

474337504 '

v — 231 v231 nv

231 + nv — Vseconds.

86. The foregoing calculations serve only for the highervelocities, such as exceed 300 or 400 feet per second of time.But, for those that arc below 400, the rule is simpler, as theresistance is then, by cor. 2 prob. 2, -0000044dV = cd’-v z ,where d denotes the diameter of any ball. Hence then,

employing the same notation as before, =

22 fx = 32x x therefore x —

C d*

and —vv --—, the correct

fluent of which is x —

32c(P

x h. 1.

87. Now, for an example, suppose the first velocity to be300 = and the last v = 100, for a 24lh ball. Thenw =24 , d — 5-546, d 1 — 30'76, c = -0000044 ; therefore

—- _ 3 r-.n __I V _ 31,0 _ -X fl,„ lii.n 7/^rv_ of

32123-04c "" „ I00

which is l-09d6; therefore 1-0986 x 5542 = 0108 = x, isthe distance. It the first velocity he only 200 = v ; then~ = 2, the hyp. log. of which is-69315, therefore -69315 x5542 = 3841 — X, tlie distance.

and

3, the hyp. log