238

THEORY AND PRACTICE

TRACT 37.

48. Ex. 5. For any other ball, whose diameter is d, andits weight w, the resistance of the air being d v *

4

152542

MV, putting b — the retarding force will be

. l * _ , v bfctP + w . . — w vv i

thence — vv — 2 gx x - , and x — — X —, and

° 9g ld*v* + u)

the cor. flu. *• =

X h. 1.

— » i 6ef , v Q + u?

4^^ *" v ** bd’ i v' 1 + u> Qfgbet 2 ^ w

for the whole height when v = 0. Now if the ball be a 24pounder, whose diameter is 5-6 nearly, and its square 31‘36;then bd‘ — -00020/14, and

4fiM

v — 2000, MV = 829, and

24

+ to

Sid 1829 + 24

= 1810; also85824 »

853

24

therefore x — 1810 x h. I. = 1810 x 3-57071 = 6463,being more than double the height of that of the small ball,or a little more than a mile, and nearly the same as in the2d example to prob. 4.

PROBLEM VI.

49. To ascertain the Time of the Ball’s ascending to theHeight determined in the last Problem, by the same Pro-jectile Velocity as there given.

By that prob. x = x ——

2 * a „+:

•, theref.i = — = '“'x

’ » 2g-a

v a +

the fluent of which, by form 11, is ~J~~ X arc to

J 1 xga * w

radius 1 tang. —^ ~ X arc tan. —or by cot-

v'~

rection t — —-\/~ x (arc tang.

— arc tang.

W

V •—

v n

),

the time in general when the first velocity v is reduced tov. And when v — 0, or the velocity ceases, this becomest = -—f-- X arc t0 tang. --for the time of the whole

2g a to

ascent; which is also equal to x arc to tang, v and rad.

v'—.

r A