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For the supplementary pile = 30 — 11 = 19, and p=30-20+1=11, 6=20-11-9; hence,

N" = 4 b (b + 1) (2 l+p) = † × 9 × 10 (38 +11) = 735, and N-N'-N" — 4235.


1. How many balls are contained in an incomplete triangular pile of 17 courses, a side of the top containing 8?

Ans. 2516. 2. Required the number of shot in an incomplete square pile of 17 courses, a side of its base containing 24.

Ans. 4760.

3. How many shells are in an incomplete rectangular pile of 12 courses, the number in the length and breadth of the base being 40 and 20? Ans. 6146.

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The subject of projectiles, considered in a practical point of view, treats of the methods of determining by calculation various circumstances belonging to the motions of bodies projected in the atmosphere.

This subject is divided into two parts: the parabolic theory of projectiles, and military projectiles or practical gunnery.


In the parabolic theory several hypotheses not strictly correct are made; but only one of them can lead to any sensible error in practice, though in some cases the error is comparatively small. This last hypothesis is that there is no resistance from the atmosphere to the motion of a projectile; and the other two are that gravity acts in parallel lines over a small extent of the earth's surface, and that its intensity is constant from its surface to a small height above it.

190. PROBLEM I.-Of the height fallen through by a body, the velocity acquired, and the time of descent, any one being given, to find the other two.



h the height fallen through,
the velocity acquired,
t the time of descent,
g=32.2 feet,

h = vt=gth =


v=gt = √2 gh=



These relations of h, v, and t, are proved in treatises of theoretical mechanics.* Any two of these three quantities are said to be due to the other; thus the velocity acquired by falling from a given height is said to be due to that height, and so of the other two quantities. The acquired velocity is also called the final velocity. The number 32.2 is the velocity in feet that a body acquires in falling during one second. The velocity with which a body is thrown upwards or downwards, is called its initial velocity.

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In solving the following exercises, such a formula is to be chosen in each case as contains the elements concerned, that is, the quantities given and sought.


1. What is the velocity acquired in falling 10 seconds? v=gt=32·2 × 10 = 322.

2. What is the height fallen through in 5 seconds?
h = gt2
= = 1 × 32·2 × 5o — 402·5.



1. What velocity would be acquired in falling 120 feet? Ans. 87-9.

2. Required the height through which a body must fall to acquire the velocity of 1500 feet per second. Ans. 34938.

* See Natural Philosophy, First Part, Chambers's Educational

Course, art. 160.

3. In what time will a body acquire the velocity of 900 feet? Ans. 27.9 seconds. 4. In how many seconds would a body fall 27000 feet? Ans. 40.9.

191. When a body is projected in any direction except that of a vertical line, it describes a parabola.

Thus, if a body is projected in the direction PT, it will describe a curvilineal path, as PVH, which will be a parabola.


192. The velocity with which the body is projected is called the velocity of projection.

The velocity in the direction of projection PT is con


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During the time that the projectile would be carried, by the velocity of projection continued uniform, to T, it would be carried by the force of gravity from T to H. But the distance PT is evidently proportional to the time, whereas TH is proportional to the square of the time. Since (190) h=gt2 = 16·12, therefore TH is proportional to the square of PT. And the same is true for any other line drawn, as TH, from a point in PT to the curve; and this is a property of the parabola (Geom. Conic Sections, Par. VII. Cor. 5.)

193. The velocity of projection is that due to a height equal to the distance of the point of projection from the directrix.

Or, the velocity at P is that acquired in falling down MP, AM being the directrix.

194. The velocity in the direction of the curve at any other point in it, is equal to the velocity due to its distance from the directrix.

The velocity at any point, as H, is that due to AH, and if a body were projected with that velocity in the direction of the tangent HG, it would describe the same curve HVP, and on arriving at P, would have the velocity due to MP.

195. The height due to the velocity of projection is called the impetus.

Thus, MP is the impetus.

196. The distance between the point of projection and any body to be struck by the projectile, is called the range, and sometimes the amplitude. When the range lies in a horizontal plane, it is called the horizontal range.

Thus, P being the point of projection, and H the body struck, PH is the range, and PQ the horizontal range.

197. The time during which a projectile is moving to the object, is called the time of flight.

198. The angle contained by the line of projection, and the horizontal plane, is called the angle of elevation.

Thus, TPQ is the angle of elevation.

199. The inclination of the plane passing through the point of projection and the object, to the horizontal plane, is called the angle of inclination.

Thus, HPQ is the angle of inclination.

The range of a projectile may be either on a horizontal or an oblique plane.

1. Projectiles on Horizontal Planes.

200. The following formulas afford rules for calculating the impetus, range, velocity of projection, time of flight, and


Leth the impetus MP in feet,


the velocity of projection in feet per second,


t = the time of flight in seconds,
the horizontal range = PH,
the angle of elevation,


r' — the greatest range,


the greatest height = VD,

v2 then h by 190; 2g

√2gh by 190;



t = 2 sin e√

r = 2h sin 2 e;


h'h sin2 e.

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Let PT be the line of projection, and PVH the curve described. On PM describe a semicircle MBP, and from its intersection with the tangent PT in B, draw BC parallel to the axis, and BA perpendicular to the impetus MP. Then (Geom. vol. II. Conic Sections, Par. XIII.) AB= PC=1PH=1r, and BC=1TH, and VDBC. Draw the radius OB, then (Pl. Geom. III. 31) angle BPC or e = M = POB, or POB = 2e. Now, OB: AB rad: sin O, or thr1: sin 2e;



hence, r=1h sin 2e, and r2h sin 2 e.





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Again, the time of flight is just equal to the time of describing PT uniformly with the velocity of projection, or the time of falling through TH by gravity. Now, if



PH: HT rad: tan TPH, or r: h"=1: tane;

h"=rtane 2h sin 2 e tan e 2h

and therefore

h"-4 h sin2 e.

But if t is the time due to h", then
2h" 8h sin2 e



2sin e cose⚫ sin e


=2 sin e√




which is the expression above for t.

Again, VDBC4TH, or h'=h"= h sin2 e.
When e 15° or 75°, sin 2 e = sin 30° = 1⁄2, and r =


The greatest value of r or 2 h sin 2 e, for a given value of h, is when sin 2e is a maximum or 2 e= 90, and e = 45. For then sin 2 e = 1, and r = =2h.

201. Two elevations, of which the one is as much greater than 45° as the other is less, give the same horizontal range.

For if these angles of elevation are 45 + d and 45 — d, then for these elevations 2 e is 90+ d and 90-d, which

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