PROJECTION. 
J3C, CD, &c. in the line AD, if it was not 
drawn continually down below that line by the 
action of gravity. Draw BE, CE, DG, &c. in 
the direction of gravity, or perpendicular to the 
horizon , and take BE, CF, DG, &c. equal to 
the spaces through which the body would de- 
scend by its gravity in the same times in which 
it would uniformly pass over the sp ices AB, 
AC, AD, &c. by the projectile motion. Then, 
since by these motions, the body is carried over 
the space AB in the same time as the space BE. 
and the space AC in the same time as the space 
CE, and the space AD in the same time as the 
space DG, &c. ; therefore, by the composition 
ot motions, at the end of those times the body 
will be found respectively in the points E, F, G, 
&c. and consequently the real path of the pro- 
jectile will be the curve line AEFG, &c. But 
the spaces AB, AC, AD, &c. being described 
by uniform motion, are as the times of descrip- 
tion ; and the spaces BE, CF, DG, &c. described 
jn the same times by the accelerating force of 
gravity, are as the squares of the times ; conse- 
quently the perpendicular descents are as the 
squares. of the spaces in AD, 
that is, - - - BE, CF, DG, &c, 
are respectively propor- 
tional to - AB 2 , AC 2 , AD 2 , &c. 
which is the same as the property of the para- 
bola. Therefore the path of the projectile is the 
parabolic line AEFG, &c. to whidi AD is a 
tangent at the point A. 
Hence, 1. The horizontal velocity of a pro- 
jectile is always the same constant quantity, in 
every point of the curve ; because the horizon- 
tal motion is in a constant ratio to the motion in 
AD, which is the uniform projectile motion ; 
viz. the constant horizontal velocity being to 
the projectile velocity, as radius to the cosine 
of the angle DAH, or angle of elevation or de- 
pression of the piece above or below the hori- 
zontal line AH. 
2. The velocity of the projectile in the direc- 
tion of the curve, or of its tangent, at any point 
A, is as the secant of its angle BAI of direction 
above the horizon. For the motion in the hori- 
zontal direction AI being constant, and A1 be- 
ing to AB as radius to the secant of the angle 
A ; therefore the motion at A, in AB, is as the 
secant of the angle A. 
3. The velocity in the direction DG of gra- 
vity, or perpendicular to the horizon, at any 
point G of the curve, is to the first uniform pro- 
jectile velocity at A, as 2GD to AD. For the 
times of describing AD and DG being equal, 
and the velocity acquired by freely descending 
through DG being such as would carry the 
body uniformly over twice DG in an equal time, 
and the spaces described with uniform motions 
being as the velocities, it follows that the space 
AD is to the space 2DG, as the projectile velo- 
city at A is to the perpendicular velocity at G. 
III. The velocity m the direction of the curve, 
at any point of it, as A, is equal to that which 
is generated by gravity in freely descending 
through a space which is equal to one-fourth of 
the parameter of the diameter to the parabola 
at that point. (Fig. 2.) 
Let PA or AB be the height due to the velo- 
city of the projectile at any point A, in the di- 
rection of the curve or tangent AC, or the ve- 
locity acquired by falling through that height ; 
and complete the parallelogram ACDB-. Then 
is CD — AB or AP, the height due to the velo- 
city in the curve at A ; and CD is also the height 
due to the perpendicular velocity at D, which 
will therefore be equal to the. former r but, by 
the last corollary, the velocity at A is to the 
perpendicular velocity at D, as AC to 2CD ; 
and as these velocities are equal, therefore AC 
or BD is equal to 2CD or 2AB; and hence AB 
or AP is equal to ijBD, or of the parameter of 
the diameter AB, by the nature of the parabola. 
Hence, L If through the point P, the line PL 
is drawn perpendicular to AP ; then the velo- 
city in the curve at every point, will be equal to 
the velocity acquired by falling through the 
perpendicular distance of the point from the 
said line PL; that is, a body falling freely 
through 
PA, acquires the velocity in the curve at A, 
EF, - - - at F, 
KD, - - - - at D, 
LH, - - - at IL 
The reason of which is, that the line PL is what 
is called the directrix of the parabola ; the pro- 
perty of which is, that the perpendicular to it, 
from every point of the curve, is equal to one- 
fourth of the parameter of the diameter at that 
point, viz. 
PA — -J the parameter of the diameter at A, 
EF — - - • - - at F, 
KD — - - - - at L, 
LH- - - - - at H. 
2. If a body, after falling through the height 
PA, which is equal to AB, and when it arrives 
at A, if its course is changed, by reflection 
from a firm plane AI, or otherwise, into any di- 
rection AC, without altering the velocity; and 
if AC is taken equal to 2AP or 2AB, and the 
parallelogram is completed; the body will de- 
scribe the parabola passing through the point D. 
3. Because- AC =: 2AB, or 2CD, or 2AP ; 
therefore AC 2 2AP 2CD, or AP . 4CD; 
and because all the perpendiculars EF, CD, GH, 
are as AE 2 , AC 2 , AG 2 therefore, also AP . 4EF 
=: AE 2 , and AP . 4GH — AG 2 , &c. ; and be- 
cause the rectangle of the extremes is equal to 
the rectangle of the means, of four proportionals, 
therefore it is always, 
ap : ae : : ae : 4ef, 
and AP ; AC ( * AC f 4CD, 
and AP ; AG *: AG * 4GH, 
and so on. 
IV. Having given, the direction of a projectile, 
and the impetus or altitude due to the first ve- 
locity ; to determine the greatest height to 
which it will rise, and the random or horizontal 
range. (Fig. 3.) 
Let AP be the height dire to the projectile 
velocity at A, or the height which a body must 
fall to acquire the same velocity as the projectile 
has in the curve at A ; also AG the direction, 
and AH the horizon. Upon AG let fall the per- 
pendicular PO, and on- AP the perpendicular 
QR; so shall AR be equal to the greatest alti- 
tude CV, and 4RO equal to the horizontal 
range AH. Or, having drawn PQ perpendicu- 
lar to AG, take AG == 4AQ, and draw GH 
perpendicular to AH; then AH is the range. 
For, by the last cor. AP * AG * * AG * 4GH, 
and by sim. triangles, AP * AG- " AQ * GH, 
or AP : AG' : : 4AQ : 4GH; 
therefore AG — 4AQ ; and, by similar triangles, 
AM =4RQ. 
Also, if V is the vertex of the parabola, then 
AB or jAG =: 2AQ, or AQ = OB; conse- . 
quently AR — BV, which i3 CV by the na- 
ture of the parabola. 
Hence, 1. Because the angle Q is a right an- 
gle, which is the angle in a semicircle : there- 
fore if upon AP as a diameter a semicircle is 
described, it will pass through the point Q. 
(Fig. 4.) 
2. Tf the horizontal range and the projectile 
velocity are given, the direction of the piece- so 
as to hit the object Id will be thus easily found: 
Take- AD =: -JAH, and draw DQ perpendicular 
to AH, meeting the semicircle described on the 
diameter AP in Q and q : then either AQ or A q 
will be the direction of the piece. And hence 
it appears, that there are two directions AB and, 
A b which, with the same projectile velocity, 
give the very same horizontal range AH : and 
these two directions make equal angles qAD and 
503 
QAP with AH and AP, because the arc PQ is 
equal to the arc Ay. 
3. Or if the range AM and direction AB are 
given, to find the altitude and velocity or im- 
petus : Take AD = -JAH, and erect the perpen- 
dicular DQ meeting AB in O ; so shall DO be 
equal to the greatest altitude CV. Also erect 
AP perpendicular to AH, and QP to AQ so 
snail AP he the height due to the velocity. 
4. When the body is projected with the same 
velocity, but in different directions , the hori- 
zontal ranges AH will be as the sines of double 
the angles of elevation; or, which is the same 
tiling, as the rectangle of the sine and cosine of 
elevation. For AD or RO, which is -|AH, is 
the sine of the arc AO, which measures double 
the angle QAO of elevation. 
And when the direction is the same, but the 
velocities different, the horizontal ranges are as 
the square of the velocities, or as the height AP, 
which is as the square of the velocity; for the 
sine AD or RQ, or jAH, is as the radius, or as 
the diameter AP. 
Therefore, when both are different, the ranges- 
are in the compound ratio of the squares of the 
velocities, and the sines of double the angles of 
elevation. 
5. The greatest range is when the angle of 
elevation is half a right angle, or 45°. For the 
double of 45 is 90°, which has the greatest sine. 
Or the radius OS, which is ^ of the range, is the 
greatest sine. 
And hence the greatest range, or that at an 
elevation of 45°, is just double the altitude AP,. 
which is due to the velocity, or equal to 4VC. 
And consequently, in that case, C is the focus 
of the parabola, and AH its parameter. 
And the ranges are equal at angles equally 
above and below 45°. 
6. When the elevation is 15°,- the double of 
which, or 30°, having its sine equal to half the 
radius, consequently its range will be equal to> 
AP, or half the greatest range at the elevation- 
of 45°; that is, the range at 15° is equal to the 
impetus or height dbe to the projectile velocity . 
7. The greatest altitude CV, being equal to 
AR, is as the versed sine of double the angle of 
elevation, and also as AP or the square of the 
velocity. Or as the square of the sine of eleva- 
tion, and the square of the velocity ; for the 
square of the sine is as the versed sine of the 
double angle. 
3.. The time of flight of the projectile, which 
is equal' to the- time of a body falling freely 
through GH or 4CV, 4 times the altitude, is 
therefore as the square root of the altitude, or 
as the projectile velocity and sine of the eleva- 
tion. 
9. And hence may be deduced the following 
set of theorems, for finding all the circumstances 
relating to projectiles on horizontal planes, hav- 
ing any two of them given. Thus, let 
s,c, t — sine, cosine, and tang, of elevation 
S,.v = sine and vers of double the elevation 
R the horizontal rage, T the time of flight, V 
the projectile velocity, II the greatest height' of 
the projectile, g — 16^ feet, and a — the im- 
petus or the altitude due to the velocity V 
Then* ' 
R — 2aS = 4asc = ~ =. = - £* 
2 g S s- t 
4h. 
H _ 
s s 
V = */ 4 ag = ^ =V— = 
T = — = 2 s V* - =V— = */— = V* 
£ £ £ S c v gf 
H = «* = \av =~ R — = * 0 L _ 
4< 4jr “ 
gT 2 
= 4 ? 
