TRACT 37.

OF GUNNERY.

233

velocity; then the rule v % ~ s becomes x 2000 =62500 feet = 11 =- miles; that is, any body, projected withthe velocity 2000 feet, rvoa'd ascend nouny 12 ur.les inheight, without resistance.

36. Corol. Because, by art. S3 Projectiles vol. 2 of theCourse , the greatest range is just double the height due tothe projectile velocity, therefore the range, at an elevationof 45°, with the velocity in the last example, would be 23Jmiles, in a nonresistmg' medium. We shall now see whatthe effects w-ill be with the resistance of the air.

PROBLEM IV.

37. To determine the Height to which a Ball, projected Up-wards as in the last Problem, will ascend, being Resisted bythe Atmosphere.

Putting x to denote any variable and increasing height as-cended by the ball; v its variable and decreasing velocitythere; d the diameter of the ball, its weight being w; >n —00000757, and n — -00175, the coefficients of the two termsdenoting the law of the air’s resistance. Then (mv x — nv)d x , by cor. 1 to prob. 2 , will be the resistance of the airagainst the ball in avoirdupois pounds; to which if theweight of the ball be added, then (mv x — nv)d x -f w will bethe whole resistance to the ball’s motion ; this divided by w,

... . (mu 3 —• nv)d* 4- w

the weight oi the bail in motion, gives--- =3

mu' 1 — nv

- d’ + l = f the retarding force, nence tile general

10 J °

formula v-0 — 2gfx (theor. 10 pa. 342 vol. 2 of the Course )

, . ' (mv 2 — n<v) d a + w . . •

IV O/-v rv, _ n a* vy '_- ■ _ m o b I /w .. • . n ... l .

becomes — v v — 2 gx Xbecause v is decreasing, where g

• _ rp «ti

*-x

making v negative16 feet; and hence

V* - V *

{mu* —nv)d' i +w Hgmdt 3

jyt 711J 3

Now, for the easier finding the fluent ofr this, assume" = then, = and,— ^ + ^+^r ;

n

3m