, PROJECTILES. 
T= v/o = v=2 n 
K 2 F V 
D — FTT = VV=: TV 
4 F 2 
F z=V = VV 
TT 2T 4D 
All which equations are very easily de- 
duced from the two original ones, D = 
F T T, and Yz=.^ FT, already demon- 
strated; the former in the proposition it- 
self, and the latter in the coiollary to it ; 
by which it appears that the nieasure.of the 
velocity at the end of the first second is 
2 F; whence the velocity (V) at the end of 
(T) seconds must consequently be expressed 
by 2F X T or 2 FT. 
Theorem 1. A projected body, whose 
line of direction is parallel to the plane of 
tlie horizon, describes by its fall a parabola, 
if the heavy body is thrown by any extrin- 
sical force, as that of a gun or the like, 
from the point A, (Plate Perspective, &c. 
fig. 7.) so that the direction of its projec- 
tion 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 direction ; and the times 
wherein the parts of space, A B, A C, A D, 
A E, were passed over, would be as the 
spaces, A B, AC, AD, &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 
constantly decline from the right line, 
AE ; and the spaces of descent, ,or the 
deviations from the horizontal line, A E, 
will be the same as if it had fallen perpen- 
dicularly. Wherefore if the body falling 
perpendicularly by the force of its gravity, 
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 AD; the 
spaces, A K, A L, AM, will be as the 
squares of the times, that is, as the squares 
of the right lines, A B, AC, AD, &c. 
orKF, LG, MH. But since the impe- 
tus in the direction parallel to the hori- 
zon always remains tire same (for the force 
of gravity, that only solicits the body 
downwards, is not in the least contrary 
to it) ; the body will be equally pro- 
moted forwards in thfe direction parallel to 
the plane of the horizon, as if there was 
no gravity at all. Wherefore, since in the 
time, AB, the body passes over a spac^ 
equal to A B ; 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 pi ojectile will be in the curve, 
A F G H I ; but because the squares of the 
right lines, K F, Ldi, M H, N I, are pro- 
portionable to the abscisses, A K, A L, 
AiM, AN. The curve, AFGHI, will 
he a semi-parabola. The path, therefore, 
of a heavy body projected according to the 
direction, A E, will be a semi-parabolical 
curve, Q E D. 
Theorem 2. Tlie curve line, that is de- 
scribed by a heavy body projected ob- 
liquely and upwards, according to any di- 
rection, is a parabola. 
I.et A F (fig. 8) be the direction 
of projection, any ways inclined to the 
horizon, gravity being supposed not to 
act, the moving body would always conti- 
nue 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 continually 
to decline from the path A F, and to move 
in a curve, which will be a parabola. Let 
lis suppose the heavy body falling perpendi- 
cularly in the time A B, through the space 
AQ, and in the time AC, through the 
space A R, &c. The spaces A Q, A R, A S, 
wilt be as the squares of the times, or as 
the squares of A B, A C, A D. It is mani- 
fest from what was demonstrated in the last 
theorem, that if in the perpendicular B O, 
there is taken B M = A Q and the paralle- 
logram 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 
deviations B M, &c. from the right lino 
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 AB, AC, 
A D. Through A draw the horizontal 
right line AP, meeting the path of the 
projectile in P. From P raise the perpen- 
dicular P E, meeting the line of direction 
in E ; and by reason the triangles A B G, 
A C H, &c. are equiangular, the squares 
of the right lines A B, A C, &c. will be pro- 
portionable 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 tlie right lines A G, AH, &c. Let 
