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(3.) Find the time of generating a velocity of 100 feet per second, and the whole space descended. Ans. 32" time, 1555 ft. space. (4.) Find the time of descending 400 feet, and the velocity at the end of that time, Ans. 4" time, 160 velocity. (5.) A body is projected downwards with a velocity of 30 feet per second: what space will it describe in 6 seconds?

Ans. 759 feet. (6.) A body is projected upwards from the top of a tower 200 feet in height, with a velocity of 40 feet per second, at the same time that another is projected upwards from the bottom of the tower with the velocity of 90 feet per second; where will they meet? Ans. 97 feet below the top of the tower. (7.) Two weights W and w weighing 8 and 5 pounds respectively, hang freely over a pulley; how far will W descend after the commencement of motion in 2 seconds? Ans. 14 feet.

(8.) Two weights W and w hang over a pulley, and W = 2 w; find the space through which a body will descend by the force of gravity, whilst W descends 2 feet.

Ans. 6 feet.

(9.) The space described by a heavy body in the 4th second, is to the space described in the last second except 4, as 1 to 3; find the whole space described. Ans. 3618 feet 9 inches. (10.) A body has fallen from A to B, when another body is let fall from C; now far will the latter body descend before it is overtaken by the former? Ans. If AB = a, and AC = b; then space descended

− (b − a)2

4 a

PROP. XV.

26. If a body be projected in free space, either parallel to the horizon, or in an oblique direction, by the force of gun-powder, or any other impulse; it will, by this motion, in conjunction with the action of gravity, describe the curve line of a parabola.

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Let the body be projected from the point A, in the direction AD, with any uniform velocity; then, in any equal portions of time, it would, therefore, describe the equal spaces AB, BC, CD, &c., in the line AD, if it were not drawn continually down below that line by the action of gravity. Draw BE, CF, DG, &c., in the direction of gravity, or perpendicular to the horizon, and equal to the spaces through which the body would descend by its gravity, in the same tîmes in which it would uniformly pass over the corresponding spaces AB, AC, AD, &c., by the projectile motion. Then, since by these two motions, the body is carried over the space AB, in the same time as over the space BE, and the space AC in the same time as the space CF, and the space AD in the same time as the space DG, &c.; therefore, by the composition of motions, at the end of those times, the body will be found respectively in the points E, F, G, &c.; and con

sequently the real path of the projectile will be the curve line AEFG, &c. But the spaces AB, AC, AD, &c., described by uniform motion, are as the times of description; and the spaces BE, CF, DG, &c., described in the same times by the accelerating force of gravity, are as the squares of the times; consequently, the perpendicular descents are as the squares of the spaces in AD, that is BE, CF, DG, &c., are respectively proportional to AB2, AC2, AD2, &c.; which is the property of the parabola. Therefore, the path of the projectile is the parabolic line AEFG, &c., to which AD is a tangent at the point A.

Corol. 1. The horizontal velocity of a projectile is always the same constant quantity, in every point of the curve; because the horizontal motion is in a constant ratio to the motion in AD, which is the uniform projectile motion. And the constant horizontal velocity, is in proportion to the projectile velocity, as radius to the cosine of the angle DAH, or angle of elevation or depression of the piece above or below the horizontal line AH.

Corol. 2. The velocity of the projectile in the direction 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 horizontal direction AI is constant, and AI is to AB, as radius to the secant of the angle A; therefore, the motion at A in AB, is every where as the secant of the angle A.

Corol. 3. The velocity in the direction DG of gravity, or perpendicular to the horizon at any point G of the curve, is to the first uniform projectile velocity at A, or point of contact of a tangent, as 2GD to AD. For, the times in 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, therefore the space AD is to the space 2DG, as the projectile velocity at A, to the perpendicular velocity at G.

PROP. XVI.

27. 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.

G

P

E

D

F

A

H

Let PA or AB be the height due to the velocity of the projectile at any point A, in the direction of the curve or tangent AC, or the velocity acquired by falling through that height; and complete the parallelogram ACDB. Then is CD = AB or AP, the 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 1⁄2BD, or of the parameter of the diameter AB.

Corol. 1. Hence it appears, 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; that 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 HF

B

TT H H H H

E

E

E

E

E

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 a firm plane AI, or otherwise, into any direction AC, without altering the velocity; and if AC be taken = 2AP or 2AB, and the parallelogram be completed; the body will describe the parabola passing through the point D.

Corol. 3. Because AC = 2AB or 2CD or 2AP, therefore AC2 = 2AP × 2CD or AP. 4CD; and because all the perpendiculars EF, CD, GH, are as AE, AC2, AG3, therefore also AP.4EF = AE2, and AP. 4GH = AG3, &c.; and, because the rectangle of the extremes is equal to the rectangle of the means of four proportionals, therefore, always

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28. 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. Upon 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 PQ perpendicular to AG, take AG = 4AQ, and draw GH perpendicular to AH; then AH is the range.

R

A

B

G

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

Also, if V be the vertex of the parabola, then AB or AG = 2AQ, or AQ= QB; consequently, AR = BV, which is = CV by the property of the parabola. Corol. 1. Because the angle Q is a right

angle, which is the angle in a semicircle, therefore, if upon AP, as a diameter, a semicircle be described, it will pass through the point Q.

Corol. 2. If the horizontal range and the projectile velocity be given, the direction of the piece, so as to hit the object H, will be thus easily found: Take AD=4AH, draw DQ perpendicular to AH, meeting the semicircle described on the

P

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diameter AP, in Q and q; then AQ or Ag 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.

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 perpendi

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cular DQ, meeting AB in Q; so shall DQ be equal to the greatest altitude CV Also, erect AP perpendicular to AH, and QP to AQ; so shall AP be the height due to the velocity.

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 4AH, 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 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.

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 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. Also, the ranges are equal, at angles equally above and below 45°.

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 45o; that is, the range at 15o, is equal to the impetus or height due to the projectile velocity.

Corol. 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 elevation, and the square of the velocity; for the square of the sine is as the versed sine of the double angle.

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.

92. SCHOLIUM. 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 e denote the elevation, R the horizontal range, t the time of flight, v the projectile velocity; h, the greatest height of the projectile; g = 32; feet, and a the impetus or altitude due to the velocity v; then

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And from any of these, the angle of direction may be found. Also, in these

theorems, may in many cases be taken = 16, without the small fraction which will be near enough for common use.

PROP. XVIII.

30. 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 the piece, and AP the altitude due to the projectile velocity at A.

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K
DC

H E

A k K

E

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 projectile pass through I, and the range on the oblique plane will be AI. For if AH, AI, be any two lines terminated at the curve, and IF, HE, parallel to the axis; then is EF parallel to the tangent AG.

31. Otherwise, without the horizontal range. Draw PQ perpendicular to AG, and QD perpendicular 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.

32. Draw Pq making the angle APq

4Aq, and draw GI perpendicular to AH.

the angle GAI; then take AG= Or, draw qk perpendicular to AH,

and take AI4 Ak. Also, kq will be equal to cv, the greatest height above

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Corol. 1. If AO be drawn perpendicular to the plane AI, and AP be bisected by the perpendicular STO; then with the centre O describing a circle through A and P, the same will also pass through 7, because the angle GA,

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