Theorem 1. A projected body, whose line of direction is parallel to the plane of the horizon, describes by its fall a pa 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. 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 ll, 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 ; Mid the spaces of descent, or the devia tions from the horizontal line, A K, 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 All ; 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, All, &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 G H I; but because the squares of the right lines, K F, L G, M H, NI, are proportionable to the abseisses, A K, AL, AM, AN; the curve, A FGH 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. rizun, 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, &e. 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 perpendicularly in the time A B, through the space A Q, and in the time A C, through the space A R, &c. The spaces A 4, 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 II 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 4, 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, A C, &c. will be proportionable to the squares of A G, A 11, &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) LXEP= APq, but APq.: AG q. ::EP:BM::LXEP:Lx BM; whence since it is L XEP=AP q. it will be L XBM= AG g. In like man ner it will be L xCN.= A fl q, &c. But because it is B G : AG:: (E P A P : by hypothesis) A P : L ; it will be L xl3G=AGXAP=AGX AG+ AGXGP=AGq.+AG X G P. But it has been shown that it is Lx BM=AG q, wherefore it will be L X BG—LxBM=AGXGP, that is, L X MG= AG X G P. By the same way of reasoning it will be L XNH=AH x H P, &c. Wherefore the rectangle under M G and L, will be equal to the square of A G, which is the property of the parabola ; and so the curve AMN 0 P K, wherein the projec tile is moved, will be a parabola.
Cor. 1. Hence the right line L is the la ths rectum or parameter of the parabola, that belongs to its axis.
Cor. 2. Let A H = H P, and it will be L x NH, whence it will be NH=CN; and consequently the right line A 1', being the line of direc tion of the projectile, will be a tangent to the parabola.
Cor. 3. It' a heavy body be projected downwards, in a direction oblique to the horizon ; the path of the projectile will be a parabola.
Theorem 3. The impetus of a projected body in different parts of the parabola, are as the portions of the tangents intercept ed between two right lines parallel to the axis ; that is, the impetus of the body projected in the points A and B (fig. 9) to which A D, and B E are tangents, will be as C D and E B, the portions of the tan gents intercepted between two right lines C B, and 1)E parallel to the axis.
We have here treated the path of a pro, jected body as an exact parabola, though, from the resistance of the air, the line of a projectile is not exactly parabolical, but rather a kind of hyperbola ; which, if con sidered and applied to practice, would render the computations far more operose, and the very small difference (as experi ence shows in heavy shot) would, in a great measure, lessen the elegancy of the demonstrations given by accounting ror it ; since the common rules are sufficient, ly exact, and easy for practice.