TRACT 37.

OF GUNNERY.

237

therefore — vv = 2gfx :and hence x = ~ x

fli4 + w

‘2sa

t 2 + --

a

the fluent of which, bv form 8, is —- x h. log. !v z + —);which when .r = 0, and -a =: v the first or projectile velocity,becomes 0 = — x h. 1. (v* + —); therefore, by subtract-

ing, the correct fluent is .r = — x h. 1. —, the height xwhen the velocity is reduced to v; and when v = 0, or thevelocity is quite exhausted, this becomes^ x h. 1. —for the whole height to which the ball will ascend.

44. Ex. 1. The values of the letters being w = l|lb,4 g — 64, a — '0009261, the last expression becomes 670 Xhyp. log. — - ———, or 1484 x com. log. ——^—. And

42880

42880

here the first vcl. v being 2000, the same expression 14S4 x

v2 x 4Wfin

log. _-2— becomes 1484 X log. of 94*28 = 2930 for the

» 42881 )

height ascended, on this hypothesis; which was 2653 by theformer problem.

45. Ex. 2. Supposing the same ball to be projected with

the velocity of only 1500 feet. Then taking 1100 velocity,whose tabular resistance is 28 - 6, being nearest to the half ofthat for 1500. Hence, as 1100* -. v z : : 28-6 : • 00002364 i/ = «m 2 .This value of a substituted in the theorem x h. 1. -

4ga V) 7

also 1500 for v, and 1^ for w, it brings out x — 2882 for theheight in this case.

46. Ex. 3. To find the height ascended by the same

ball, projected with 820 feet velocity. Here takmg 600,whose resistance 6"90 is a near medium ; then as 600 2 ; 6‘90: 1 : -0000194 = a. Hence — X h. I. = 1802 the

... ° w

height in this case.

47. Ex. 4. With the same ball, and 1640 velocity. As-sume 1200, whose resistance 35-275 is nearly a medium.Then as 1200 1 : 35-275 1 :: i : -0000245 = a. Hence — x

t>4<z

li. 1. av — 2500, the height in this case.