TRACT 37.

OP GUNNERY.

233

velocity; then the rule — s becomes -j' + - x 2000" =

62500 feet = 11£ miles; that is, any body, projected withthe velocity 2000 feet, would ascend nearly 12 miles inheight, without resistance.

36. Corol. Because, by art. 88 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 23-Jmiles, in a nonresisting medium. We shall now see whatthe effects will be with the resistance of the air.

1

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; m =00000157, and n = -00175, the coefficients of the two termsdenoting the law of the air’s resistance. Then (mv z — nv)d z , 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 — nv)d z -f- w will bethe whole resistance to the ball’s motion ; this divided bv w,

, . , i iii* • • _ + W

the weight or the ball in motion, gives-- =

mv d' 1 + 1 —f the retarding force. Hence the generalformula vv = 2gfx (theor. 10 pa. 342 vol. 2 of the Course)

. • . — ??.*u)rf 2 + w . . •

becomes — w =: c 2gx X --, making, v negative

because v is decreasing, where g = 16 feet; and hence

W VV ~ W VV

{viv‘ 1 — 7iv)d 2 + w %gmd 2

ind 2

Now, for the easier finding the fluent of this, assume

v — — — z ; then »=:?+—.

2m ’ 2m'

and v % zz z z +

and vv — zz + —z, and v* - v +

2m ’ m

and