TRACT 37.

OF GUNNERY.

241

56. For Example. With the l|!b ball of 2 inches dia-meter, and 2000 feet velocity; that is 3>:=l-§- or 1 • 125, d = 2,v = 2000; also m = -00000757, n = -00175; then — =

231, and £- = 115£ ~p, and = 37153 p 1 + q 2 , andq = \/ (37153—p z ) =z 154]-; therefore - g ^ -——7-522. Also

= 12-21, the tangent to the arc of 85° 19', the length

of which is 1-4S90. Then 7-522 + 1-489 ; 11-199 seconds,

the time of ascending by this rule. And when the velocity

v is infinite, then the tangent —^ hecomesinfinite also ; con-, . 9

secjnently its arc is a quadrant T570, and therefore 7'522 x1-57 = 3T81 seconds, is the time of ascent with an infinitevelocity.

57. Exam. 2. For any other weight of ball, as supposethe 24lb = w, the diameter being 5-6, or more nearly 5 - 546= cl. Here then d 2 = 30'758, — = -78029, — 103078

7 d 2 7 md*

— P 2 -\-q Z i P ~ 115-J as before, q— y( 103078 — p 2 ) = 299*4}= 10-76. If the velocity v 2 = 2000 as be-

_ p a +

fore, then the tangent —- — 6'293, the arc of which con-

9

tains 80° 58, the length of which is 1-413; then 10 - 76 x1-413= 15-20 seconds, is the whole time of ascent whenprojected with 2000 feet velocity. Also 10-76 x 1-570 =1G-S9 seconds, is the time when projected with an infinitevelocity.

PROBLEM VIII.

.58. To determine the same as in Prob. v, taking into the ac-count the Decrease of Density in the Dir, as the Ball asqendsin the Atmosphere.

In the preceding problems, relating to the height and timeof balls ascending in the atmosphere, the decrease of densityin the upper parts of it has been neglected, the whole heightascended by the ball being supposed in air of the same den-sity as at the earth’s surface. But it is well known that theVOL. XU. K