S24

MISCELLANEOUS

TRACT 38.

PROBLEM V.

A Ball of Lead, of 4 inches diameter, being dropped fromthe top of a tower, of 65 yards high, falls into a cistern full ofwater at the bottom of the tower, of 20-'- yards deep: it is re-quired to determine the times of falling, both to the surfaceand to the bottom of the water.

The fall in air is 195 feet, and in water 60| feet. By thecommon rules of descent, as a/ 16 : f 195 : : l" : 195 =

3-49 seconds, the time of descending in air. And as fl6 :a/195 : : 32 : 8 X /195= 11 T71 feet, the velocity at the end#f that time, or with which the ball enters the water.

'Zblead, n

Again, by prob.22 of vol. 2, art. 2 of the Course, the space

3 = Tb x hyp - log ’ of 7=5’ or rather Tb * h >’P- log ' ° f

*f~ ■■ (the velocity being decreasing, and e 2 greater than a) —

g'i _ Q f

X com. log. of where n *= 11325 the density of

256f/fN —w) L 3 n

Jk » ° — Hh> e =

11T71 the velocity at entering the water, and v the velocityat any time afterwards, also d the diameter of the ball zz 4inches, and m = 2'302585 the hyp. log. of 10.

Hence then n zz 1 1325, n = 1000, N — n = 10325, d zz— zz —; then a = - - -zz —=-r— zz 293 a, and b —

it) 'J > ‘An 9U()0

Also e — 111*71 ;

12 3

3», 9n _ 9000

SriN

3ft

15 1

8n — SOuOO — 151 10 near tv'.

. €*—-O

therefore s = 604 = — x log. of -— = 5m x log.-,

This theorem will give s when v is given, and by revertingit will give v in terms of 5 in the following manner.

Dividing by 5m gives ~ = log. of 7—‘ = ns, by putting

n = 2 -; therefore, the natural number is 10 ,J = - ~ a — .5m *» — a ’

hence v l - a = and v = V{a + which, by

substituting the numbers above mentioned for the letters,