seending 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, therefore the space AD is to the space 2DG, as the projectile velocity at A, to the perpendicular velocity at G.

PROPOSITION XX.

79. The Velocity in 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 of the Parabola at that Point.

height due to the velocity in the curve at A; and CD is also the height due to the perpendicular velocity at D, which must be equal to the former; but by the last corol. 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 BD, or of the parameter of the diameter AB, by corol. to theor. 13 of the Parabola.

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80. Corol. 1. Hence, and from cor. 2, theor. 13 of the Parabola, it appears that, if from the directrix of the parabola which is the path of the projectile, several lines He be drawn perpendicular to the directrix, or parallel to the axis; then the velocity of the projectile in the direction of the curve, at any point e, is always equal to the velocity acquired by a body falling freely through the perpendicular line HE.

81. Corol. 2. If a body, after falling through the height PA (last fig. but one), which is equal to AB, and when it arrives at A, have its course changed, by reflection from an elastic plane AI, or otherwise, into any direction Ac, without altering the velocity; and if ac be taken = 2AP or 2AB,

and

and the parallelogram be completed; then the body will describe the parabola passing through the point D.

82. Corol. 3. Because AC = 2AB or 2CD or 2AP, therefore AC2 = 2AP X 2CD or AP. 4cD; and, because all the perpendiculars EF, CD, GH, are as AE, AC, AG2; therefore also AP. 4EF = AE, and AP. 4GH = AG2, &c; and, because the rectangle of the extremes is equal to the rectangle of the means of four proportionals, therefore always

it is AP: AE :: AE : 4EF,
and AP: AC :: AC : 4CD,
and AP: AG :: AG: 4GH,
and so on.

PROPOSITION XXI.

83. Having given the Direction, and the Impetus, or Altitude due to the First Velocity of a Projectile; to determine the Greatest Height to which it will rise, and the Random or Horizontal Range.

LET AP be the height due to the projectile velocity at A, AG the direction, and AH the horizon. On AG let fall the perpendicular PQ, and on AP the perpendicular QR; So shall Ar be equal to the greatest altitude cv, and 4QR equal to the horizontal range AH. Or, having drawn

G

P

B

V

R

A

PQ perp. to AG, take AG = 4AQ, and draw GH perp. to AH;

then AH is the range.

AP: AG:: AG: 4GH;
AP AG:: AQ : GH,
AP: AG:: 4AQ: 4GH;

For, by the last corollary,

and, by similar triangles,

therefore AG = 4AQ; and, by similar triangles, AH = 4 Q

Also, if v be the vertex of the parabola, then AB or LAG = 2AQ, or AQ = QB; consequently AR = BV, which is = cv by the property of the parabola.

84. Corol. 1. Because the angle Q is a right angle, which is the angle in a semicircle, therefore if, on AP as a diameter, a semicircle be described, it will pass through the point Q.

B

P

Q

V

R

g

S

the direction of the piece so as to hit the object H, will be thus easily found: Take AD = AH, draw De perpendicular to AH, meeting the semicircle, described on the diameter AP, inq and q; then AQ or Aq will be the direction of the piece. And hence it appears, that there are two directions AB, ab, which, with the same projectile velocity, give the very same horizontal range AH. And these two directions make equal angles qAD, QAP with AH and AP, because the arc PQ = the arc Aq.

86. Corol. 3. Or, if the range AH, and direction AB, be given; to find the altitude and velocity or impetus. Take AD = AH, and erect the perpendicular DQ, meeting AB in Q; so shall Do be equal to the greatest altitude cv. Also, erect AP perpendicular to AH, and or to AQ; so shall ar be the height due to the velocity.

87. Corol. 4. When the body is projected with the same velocity, but in different directions: the horizontal ranges AH will be as the sines of double the angles of elevation.Or, which is the same, as the rectangle of the sine and cosine of elevation. For AD or RQ, which is AH, is the sine of the arc AQ, which measures double the angle QAD of elevation.

And when the direction is the same, but the velocities different; the horizontal ranges are as the square are of the velocities, or as the height AP, which is as the square of the velocity; for the sine AD Or RQ or AH 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.

88. Corol. 5. The greatest range is when the angle of elevation is 45°, or half a right angle; for the double of 45 is 90, which has the greatest sine. Or the radius os, which is

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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. Consequently, in that case, c is the focus of the parabola, and AH its parameter. Also, the ranges are equal, at angles equally above and below 45o.

89. Corol. 6. When the elevation is 15°, the double of which, or 30°, has its sine equal to half the radius; consequently then its range will be equal to AP, or half the greatest range at the elevation of 45%; that is, the range at 15o, is equal to the impetus or height due to the projectile velocity.

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90. Corol. 7. The greatest altitude cv, being equal to AR is as the versed sine of double the angle of elevation, and also as ar or the square of the velocity. Or as the square of the sine of elevation, and the square of the velocity; for the square of the sine is as the versed sine of the double angle.

91. Corol. 8. The time of flight of the projectile, which is equal to the time of a body falling freely through GH or 4cv, four times the altitude, is therefore as the square root of the altitude, or as the projectile velocity and sine of the

elevation.

SCHOLIUM.

projectile g = 16+ feet, and a the impetus, or the altitude

92. From the last proposition, and its corollaries, may be deduced the following set of theorems, for finding all the circumstances of projectiles on horizontal planes, having any two of them given. Thus, let s, c, t denote the sine, cosine, and tangent of elevation; s, v the sine and versed sine of the double elevation; R the horizontal range; r the time of flight; v the projectile velocity; H the greatest height of the

due to the velocity v. Then,

And from any of these, the angle of direction may be found. Also, in these theorems, g may, in many cases, be taken = 16, without the small fraction Tz, which will be near enough for common use.

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PROPOSITION XXII.

93. To determine the Range on an Oblique Plane; having given the Impetus or Velocity, and the Angle of Direction.

LET AE be the oblique plane, at a given angle, either above or below the horizontal plane AH; AG the direction

of

of the piece, and Ap the altitude due to the projectile velocity at A.

G

P

By the last proposition, find.. the horizontal range AH to the given velocity and direction; draw He perpendicular to AH, meeting the oblique plane in E; draw EF parallel to AG, and FI parallel to HE; so shall the

A

RK

L

projectile pass through I, and the range on the oblique plane will be AI. As is evident by theor. 15 of the Parabola, where it is proved, that if AH, AI be any two lines ter minated at the curve, and IF, HE parallel to the axis; then is EF parallel to the tangent AG.

94. Otherwise, without the Horizontal Range.

Draw PQ perp. to AG, and QD perp. to the horizontal plane AF, meeting the inclined plane in K; take AE = 4AK, draw EF parallel to AG, and FI parallel to AP or DQ; so shall AI be the range on the oblique plane. For AH = 4AD, therefore EH is parallel to FI, and so on, as above.

Otherwise.

95. Draw pq making the angle Arq = the angle GAL; then take AG = 4Aq, and draw GI perp. to AH. Or, draw qk perp. to AH, and take AI = 4Ak. Also kq will be equal

to cu the greatest height above the plane.

therefore AG = 4Aq; and by sim. triangles, AI = 4ak.

Also, qk, or GI, is = to cv by theor. 13 of the Parabola.

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96. Corol. 1. If Ao be drawn perp. to the plane AT, and AP be