252

THEORY AND PRACTICE

TRACT 37.

80. Now, for an application, let it be required first, todetermine in what space! a 24ll> ball will have its velocity re-duced from 1780 feet to 1500, that is, losing 280 feet of itsfirst velocity. Here d — 5-546, w = 24, v = 1780, and

1500: also — =231.’ Hence x = 7420 x loir. _ 1 =

m ” v — 231

1549

7420‘4 x log. J2 ~^ = 642 feet, the space passed over whenthe ball has lost 280 feet of its motion.

81. Again, to find with what velocity the same ball willmove, after having described 1000 feet in its flight. The

above theorem is ,v or 1000 = 7420 x log. v ~ or

= the common log. of ; but the number to the com.

o t—231 7

7 1000 . , 1549 ,

,0 o- 7421) 1S 1-3635 ~ N Sl, l 1 P OSe 5 then N ^ JZTiiT» an(1 Nv ~

23 In = 1549, or n» = 1549 4- 231n, and v = 4- 231 =

1 136 + 231 = 1367, the velocity when the ball has moved1000 feet.

82. Next, to find a theorem for the time of describingany space, or destroying any velocity.

Here t

_ IV

3 Snui* ^

3

the fluent

of which, by the 9th form, is / =

TV 1 t i V

——r, X — X h. 1. -

32md s q v — q

JZ- x i, i. -» ■

-, and by correction

t =

x (h. 1. -

N V

JL — l3 * sd

v —231* V

HXmfPq

- _ i„ l. —) = —’V

7 v — 7 *}xmU' 2 q

X hyp. log.

„x com. log. ^72?. —, putting v for the first

2;u ° »— i] v 1 °

velocity, and 231 for — or a its value, as before.

m *

S3. Now, to take for an example tbe same 24lb ball, andits projected velocity 1780, as before; let it be required tofind in what time this velocity will be reduced to 786. Herethen v = 1780, v == 786, w = 24, d = 5-546, d 1 = 30-76 ;

hence

1338d

7420~231"

= 32|; and

v — 23i

1549

786

__ — TTT" X f-en

231 231 8 ’ v - 231 * v 555 ~ 1^80

1-232, the log. of which is -0906 ; then 32| x '0906 = 2"’9,tlic time required, being almost 3 seconds.