PROJECTILES. 



rabola. If the heavy body is thrown by 

 any extrinsical force, as that of a gun, or 

 the like, from the point A, (Plate Per- 

 spective, &c. fig 1 . 7.) so that the direction 

 of its projection is the horizontal line, 

 A D, the path of this heavy body will be 

 a semi-parabola. For if the air did not 

 resist it, nor was it acted on by its gravity, 

 the projectile would proceed with an 

 equable motion, always in the same di- 

 rection ; and the times wherein the parts 

 of space, A B, A C, A D, A E, were pass- 

 ed over, would be as the spaces A B, 

 A C, A D, &c. respectively. Now if the 

 force of gravity is supposed to take place, 

 and to act in the same tenour, as if the 

 heavy body were not impelled by any 

 extrinsical force, that body would con- 

 stantly decline from the right line, A E ; 

 and the spaces of descent, or the devia- 

 tions from the horizontal line, A E, will 

 be the same as if it had fallen perpendi- 

 cularly. Wherefore, if the body, falling 

 perpendicularly by the force of its gra- 

 vity, passed over the space A K in the 

 time A B descended through A L, in the 

 time A C, and through A M in the time 

 A D ; the spaces, A K, A L, A M, will be 

 as the squares of the times, that is, as the 

 squares of the right lines, A B, A C, 

 A D, &c. or K F, L G, M H. But since 

 the impetus in the direction parallel to 

 the horizon always remains the same 

 (for the force of gravity, that only soli- 

 cits the body downwards, is not in the 

 least contrary to it) ; the body will be 

 equally promoted forwards in the direc- 

 tion parallel to the plane of the horizon, 

 as if there was no gravity at all. Where- 

 fore, since in the time, A B, the body 

 passes over a space equal to AB; but 

 being compelled by the force of gravity, 

 it declines from "the right line, A B, 

 through a space equal to A K ; and B F 

 being equal and parallel to A K, at the 

 end of the time, A B, the body will be in 

 F, so in the same manner, at the end of 

 the time A E, the body will be in I ; and 

 the path of the projectile will be in the 

 curve A F GH I ; but because the squares 

 of the right lines, K F, L G, M H, NI, 

 are proportionable to the abscisses, A K, 

 A L, A M, A N ; the curve, A F G H I, 

 will be a semi-parabola. The path, there- 

 fore, of a heavy body, projected accord- 

 ing to the direction, A E, will be a semi- 

 parabolical curve, Q E D. 



Theorem 2. The curve line, that is 

 described by a heavy body projected ob- 

 liquely and upwards, according to any 

 direction, is a parabola. 

 Let A F (fig. 8) be the direction of 



projection, any ways inclined to the ho- 

 rizon, gravity being supposed not to act, 

 the moving body would always continue 

 its motion in the same right line, and 

 would describe the spaces A B, A C, A D, 

 &.c. proportional to the times. But by 

 the action of gravity it is compelled con- 

 tinually to decline from the path AF, 

 and to move in a curve, which will be a 

 parabola. Let us suppose the heavy body 

 falling perpendicularly in tiie time A B, 

 through the space A Q, and in the time 

 A C, through the space A K, &.C. The 

 spaces A Q, A R, A S, will be as the 

 squares of the times, or as the squares of 

 A B, A C, A D. It is manifest, from what 

 was demonstrated in the last theorem, 

 that if in the perpendicular B G, there is 

 taken B M = A Q, and the parallelogram 

 be completed, the place of the heavy 

 body at the end of the time A B, will be 

 M, and so of the rest ; and all the devia- 

 tions B M, &c. from the right line A F, 

 arising from the times, will be equal to 

 the spaces A Q, A R, A S, which are as 

 the squares of the right lines A B, A C, 

 A D. Through A draw the horizontal 

 right line AP, meeting the path of the 

 projectile in P. From P raise the per- 

 pendicular P E, meeting the line of di- 

 rection in E ; and by reason the triangles 

 A B G, A C II, &c. are equiangular, the 

 squares of the right lines A B, AC, &c. 

 will be proportionable to the squares of 

 A G, A H, &c. so that the deviations B M, 

 C N, &c. will be proportionable to the 

 squares of the right lines A G, A H, &c. 

 Let the line L be a third proportional to 

 E P and A P ; and it will be (by 17 El. 6) 

 L X E P = A P f/, but A P q. : A G q. 

 : : E P : B M : : L X E P : L X B M ; 

 whence since it is L X E P = A P q. it 

 will be L x B M = A G q. In like man- 

 ner it will be L X C N = A H q, 8tc. 

 But because it is B G : A G : : (E P : 

 A P : : by hypothesis) A P : L ; it will be 

 L XBG = AGXAP=AGX 

 AG -f AGxGP=AGgr. + AG 

 X G P. But it has been shown that it is 

 LxBM = AG<7, wherefore it will be 

 L X BG LxBM = AGxGP, 

 that is, L x M G = A G X G P. By 

 the same way of reasoning it will be L 

 X N H = A H X H P, 8cc. Wherefore 

 the rectangle under M G and L, will be 

 equal to the square of AG, which is the 

 property of the parabola; and so the 

 curve A M N O P K, wherein the projec- 

 tile is moved, will be a parabola. 



Cor. 1. Hence the right line L is the la- 

 tus rectum or parameter of the parabola, 

 that belongs to its axis, 



