to the horizon; as in the projection of balls and shells from mortars, or other pieces of ordnance.

189. Let a body be projected from A, in any direction, not vertical as

D AD; and let AC, AD, be the spaces that the body would describe in the times t and t', from the uniform veJucity of projection, and CE and DB the spaces through which it would de

C scend by the action of gravity in the

B same times; then it is obvious, from the composition of the two motions that the body will be found at the ends of those times in the points E and B.

Now, by the laws of uniform motions (art. 26) AC=tv, and AD= thu (v being the velocity of projection), therefore,

E AC: AD :: 1:ť; and, by the laws of falling bodies(41), sat", and therefore CE : DB :: 1 : t'!;

F whence CE : DB :: ACS : ADP, which is a known property of the parabola ; and, as the same las place for every point in the path of the projectile, it follows, that the curte described by the projectile is a parabola.

190. In order now to investigate the laws of the bodies' motion,

let A be the point of projection AB, or AB', the plane on which the body is projected, passing through A, and which also denotes the range.

Let AC be drawn parallel, and BCD perpendicular to the horizon;

or

cos. B

cos. B

:

or

=g

let the angle of elevation CAD3 A, and the angle of inclination of the plane CAB=B, the velocity of projection =v, the time of flight = t, and the range AB = r, also let16į teet = g.

Then we know from the laws of uniform motion, that the body at the end of the time t, if gravity did not act, would be found in the point D; while by the laws of falling bodies, it would in the same time pass through the perpendicular DB ; consequently

AD=tv; and DB=g18 Whence we have sin. LABD : sin. <BAD :: AD: DB

tv sin. (A+B) B : sin. (A + B) :: tv : and therefore gl_sin. (A+B)

(Equation I.) Again sin. ZADB : sin. <BAD :: AB : DB

q sin. (A + B)
cos. A : sin. (A + B) ::
and therefore gle_sin. (A + B)
COS. A

(Equation II.) And equating the value t in equations 1 and 2, we obtain sin. (A + B) cos. A

(Equation 117.) From these three equations, all the relations between the time, velocity, range, and angle of elevation, are readily determined. For example:

If the time and elevation be given to find the velocity and range ; in this case, Equation 1 gives va

sin. (A+B) Equation 2 gives r=

gte cos. A

sin. (A+B) If the range and elevation be given to find the time and velocity we have from

sin. (A+B) Equation 2

А

COS, A

T

v?

& cos. B

gt cos. B

Equation 3 If the velocity and elevation be given, to find the time and range, we obtain from

v sin. (A+B) By Equation 1...,..t=

If any iwo of the above quantities are given to find the angle elevation we must substitute, instead of sin. (A+B), its value : viz.

sin. (A+B)=sin. A cos. B+ sin. B cos. A

vi hence sin. A, or cos. A may be obtained. In all t eie cases, we shall find sin. (A+B) cos. A equal to a known quantity, which let be denoted by C, then

cos. A sin. A cos. B+sin. B cos. ?A=C
Let sin. A=x, then cos A=v(1-x2); and we have

С
*V(1-)+tang. B(I-1)=

A quadratic equation from which two values of r will always be determined ; and whence we learn that there are always two angles of elevation, which equally answer the conditions of the Problem.

191. To tind the greatest height of the projectile above the point of projection, we must observe that the body will continue to ascend till the velociiy of descent from gravity, is equal to the uniform velocity of ascent from projection ; that is, calling the time at which the height i; greatest t', we sl all have (41) 2gt'=the velocity of descent from gravity; and v sin. A will denote the uniform velocity of ascent estimated in a direction perpendicular to he horizon, whence we have

2gť=v sin. A, or

v sin. A is

2 €

bi?t the descent in the tinie i' is gt'e, in v hic , substituting the above value of 1', we have

sin.'A

48 and the ascent in the same time from projection, is t'v sin. A, or substiuting for t' as abı ve, we have

gt'l=

and consequentiy, he difference of these will be the greatest height of the projectile above the point A ; calling therefore the greatest height h, we have

А
ha
4 g

(Equation IV.

or

4

po =sin. A

whence, the angle of elevation and the velocity being given, the greatest height will be immediately determined ; and if r or l be given, ihe value of v2 may first be determined from the proper equation, and then the value of h from equation (IV.)

192. All the preceding equations are rendered much more simple if we suppose ihe plane AB to become horizontal For then the angle B=0, and consequently sin. B=0, and cos. B=i. Atier this reduction,

which formulas involve all the conditions of a projectile while the plane is horizontal, and passes through the point of projection.

When the plane is not borizontal, it is obvious chat in the formula in which sin (A + B) occurs, B must be accounteil positive when the plane descends, and negative when A ascends.

193. "If the velocity of the projectile in any point of the curve be required, or the velocity after any time 1', it' is olyvious tliat this is compounded of the constant horizontal velocity of projection v cos. A, and the difference of the vertical velocities of ascent from projection =v sin. A, id that of descent from gravity =2gt', that is, calling the required velocity V, we shall have V=cos. A+ (v sin. A-2g'1) (Equation V.)

Examples on a Horizontal Plane. 1. Let the angle in which a body is projected by 45°, and the time of its flight 12" ; what is the horizontal range? By article, 192, we have =tan. A

*

gl?

1611X122 15 ia x122 whence is

=2318'4 feet. tan. A

tan. 45° 2. In what time will a shell range 3250 feet, in an elevation of 32'.

3. If the elevation of the piece de 30o and the horizontal range 2000, what is the greatest height to which the ball will ascend ?

)

By article (192) 4kg –sin. A

4

ve sin. A whence h=

4g Now by the same articlei= sin. 2 A, or vi=

2gr

sin. 2 Asin. A cos. A and substituting this value, we obtain

h=. =7 .

-tan. A=288°7 feet.

Example on an Inclined Plane. How far will a shot range on a plane which descends 8° 15', the projectile velocity being 440 feet per second, and the elevation of the piece 32° 30' ?

By Equation III., (article 190) we have
sin. (A + B) cos. A v? S sin. (A + B) cos. A

{*
g

cos.? B
whence r=
(440)? , sin. 40° 45' x cos. 32° 30'

cos. 8° 15'
Log. sin. 40° 45'

= 9814753 cos. 32° 30'

= 9.926029 2 log. 440...

= 5:286906 2 log. cos. 8° 15' = 19*989832

25 027688 log. 107 = 1206376

.

2

& cos.B

A} 30}

160* {

log. range

= 3.831480 whence the range is 6784 feet nearly. The learner will observe, that in the preceding reasonings, no notice has been takıd of the effects of the resistance of the air on the motion of projectiles. Now, as this is very considerable, especially when they are discharged with great velocity, the theory requires to be modelled and corrected by ex. perimental investigations, before it can be applied in practice. There are indeed some cases, such as in the throwing of shells, when the velocity does not exceed 400 feet per second, in which the results by the theory do not differ much from the truth. But when the velocity is great, the resistance of the air occasions a diminution of motion so prodigions, as to revder the theory, withe ont the aid of data derived from experiment, of very little use. Thus, a musket ball, discharged with the ordinary allotment of powder, issnes from the piece with the velocity of 1670 feet per second. At the elevation of 45°, it should therefore range 16 miles, whereas it does not range above balf a mile. Thus, alsn, a 24 lb. ball, discharged with 16 lbs of powder, which should range about 16 miles, does not range 3 miles.

Again, the path of a projectile, when the velocity is great, is not paraboKcal: bot is macl less incurvaled in the neending tbau in the descending brma.