240

THEORY AND PRACTICE

TRACT 37 -

l^lb ball, wlicn w = 1|, or y'w = 1-06, this time 9'Gfw, is10'176 seconds.

54. Corol. 3. For any other size of ball, as that of d di-ameter, instead of the one of 2 inches diameter, employedabove, we must take, in the theorem, \Pa instead of a, andthen the whole time in this general case will be %■ X — V—

° d 4g- a

2 19*2

— — x 9’6fw = -j-fw. But w, the ball’s weight,

is as d } , the cube of the diameter, therefore is asor ns y'rf; that is, the whole time, for the ascent of differentballs, is as v'd, the square-root of the diameter. Thus, fora 24 pound ball, having its diameter almost 5'6, the ratio iss/ 2 to \/5’6, or 1 to ^2-8 = 1-6733 ; then 1-6733 X 10-176= 17 second's nearly, the whole time of the 24 pound ball’sascent, when projected with an infinite velocity.

PROBLEM VII.

55. To determine the Time of a Ball’s ascending to its great-est Height , using the same Formula of Resistance as inProb. 4.

Now, as in that problem, x --— x -—- =

1 2g (mu a — nv) d* + w

1 ~T * ITT ' * ? tilen dividing by v , or z + p, ~ - t =

a — — x ; the general fluent of which is t — -—

x arc to tang ^ or v —~ and radius 1; or, by correction, t =

dfTfH x (arc to tlin g- ~~ ~ arc t0 tan S- ^7^)- But > whenthe first velocity v is great, the arc to this latter tangent ”-~ p -

4 i ^ q

may be omitted, as equal to nothing, or as of no effect, sincethe value of p is 1154-, and when the small velocity v is about100 or 200, the resistance, by the formula -0000302 &v z -‘001v, here used, comes out either nothing, ora small quan-tity negative. The latter arc being rejected then, there re-mains only -*-- - X arc to tangent V -T£ to radius 1.