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are each other's supplements; and hence sin (90+ d) = sin (90―d), and the two values of r are

r = 2h sin (90+ d), and r = 2 h sin (90 — d),

which are equal.

202. PROBLEM II.-Given the velocity of projection, or the impetus and the elevation, to find the range, the time of flight, and the greatest altitude of the projectile.

v2

The formulas to be used are h= r = 2h sin 2 e,

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2g'

sin e, and h'h sin2 e.

Or r, t, and h', may sometimes be more easily found by logarithms, thus:

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Some of these rules may also be given in the form of a proportion; thus,

Radius: sin 2 e = 2 h:r; or,

Radius is to the sine of twice the elevation, as twice the impetus to the range.'

=

For (fig. to 200) angle BPC PMB, and POB = 2 M2e, and in triangle AOB

Rad: sin O=OB: AB=40B: 4 AB. Or, rad: sin 2 e 2 h: r, for 4 OB = 2 h, and 4 AB = r.

EXAMPLE.-A ball was discharged with a velocity of 1500 feet at an elevation of 24° 36'; required the range, the time of flight, and the greatest altitude.

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Lr

=

L 69876+ L sin 49° 12′ — 10 — 4·8443343 + 9.8790930-104.7234273 = L 52896.

And hh sin2 e = 34938 x 416280826054.4.
Or,
Lh+2 L sin e- 204·5432980 +

Lh'

2 x 9.6193864-203-7820708 L 6054-4.

EXERCISES.

1. A shell being discharged at an elevation of 28° 30', and with a velocity of 1610 feet in a second; what is the impetus, the range, the time of flight, and the greatest elevation? Ans. h=40250, r = 67513, h' =9164,

and t 47.7 seconds.

=

2. The impetus with which a cannon ball is fired is 3600, and the elevation 75°, and the elevation of another fired with the same impetus was 15°; required the ranges. Ans. 3600.

3. Required the time of flight of a shell fired at an elevation of 32°, with an impetus of 1808 feet. Ans. 11.23".

203. PROBLEM III.-Given the range and elevation, to find the velocity of projection.

From r2h sin 2e is found h=

r

2 sin 2e'

the 1st for

v2

mula; and from h =

is derived v√2gh, the 2d for

2g

mula.

The first gives also sin 2 e: radius=r: 2h; or,

The sine of twice the elevation is to radius, as the range to twice the impetus.'

This rule is proved by means of triangle AOB, as in art. 202; also by logarithms :

L2h 10+ Lr-L sin 2 e; and,

=

Lo=(L2g+Lh).

The greatest altitude and time of flight are found as in last problem.

EXAMPLE. A ball was projected at an elevation of 54° 20′, and was found to range 2000 feet; required the initial velocity.

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and 2gh√2 × 32·2×1055·5√67974-2260·7.

EXERCISES.

1. A shell projected from a mortar at an elevation of 60°, was found to range 3520 feet; required the impetus and velocity of projection. Ans. h= 2032, and v =367·7.

2. A ball projected at an elevation of 15° or 75° was found to range over 5200 feet; what was the impetus and velocity of discharge? Ans. h5200, and v = = 579.

3. The elevation being 45° and range 12000, what is the impetus ? Ans. 6000.

204. PROBLEM IV.-Given the impetus or projectile velocity and the range, to find the elevation.

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and r 2 h sin 2 e, therefore sin 2 e=

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By the former, 2 h: r = radius: sin 2 e; or,

'Twice the impetus is to the range, as radius to the sine of twice the elevation."

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1. At what elevation must a piece of ordnance be fired, so as to throw a ball 5600 feet, the initial velocity being

800 feet?

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e = 83° 40′ 56′′,

Hence, e6° 19′ 4′′, and 90°

which are the two elevations.

The greatest height, and the time of flight, can now be

found as in the first problem.

2. At what elevation will a mark at the distance of 5100 yards be hit with an impetus of 3000 yards.

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1. At what elevation must a shell be fired, with a velocity of 420 feet, so as to range 5400 feet?

Ans. 40° 9', or 49° 51'. 2. Required the elevation necessary to hit an object 4200 yards distant, with an impetus of 4000 yards.

Ans. 15° 50', or 74° 10′.

205. PROBLEM V. Given the elevation and time of flight, to find the range and velocity of projection.

The formulas are rgt2 cot e, v =
Or,

gt

2 sin e

L2r Ig+2 Lt + Lcote-10,
L2vLg+Lt + 10-L sin e.

EXAMPLE. A ball projected at an angle of 32° 20' struck the horizontal plane 5 seconds after; what was the range and projectile velocity?

r=gt2 cote = × 32·2 × 25 × 1·5798=636,

and v=

161

=

= 150.5;

gt 32.2 × 5
2 sin e 2 x 5348 1.0696

=

as is known, h can now be found by article 190.

v2

The formulas are obtained thus:. -Since h= and

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2g'

and therefore (200)

•sin 2e; but sin 2e = 2 sin e⋅ cos e,

=cote; hence, r = gt2 cot e.

9212

r = √2gh = √ 4 sin2 e

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EXERCISE. The time of flight of a shell projected at an elevation of 60° was 25 seconds; what was the initial velocity and the range? Ans. r5807, and v = 465.

206. Besides the preceding theorems for projectiles on horizontal planes, many more might be given of less importance; the two following are sometimes useful :—

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207. The ranges are proportional to the sines of twice the angles of elevation."

Let r and r' be two ranges corresponding to the elevations e and e', then

r:r' sin 2 e: sin 2 e'; and therefore r' —r

sin 2 e'

sin 2 e

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1. If a shell range 1000 yards at an elevation of 45°, how far will it range at an elevation of 30° 16'?

Ans. 871 yards. 2. If the range of a shell at an elevation of 45° is 3750, what must be the elevation for a range of 2810 feet? Ans. 24° 16', or 65° 44'.

3. A shell discharged at an elevation of 25° 12′ ranges 3500 feet; how far will it range at an elevation of 36° 15'? Ans. 4332.

208. 'The ranges are proportional to the impetus, or to the squares of the velocities."

Or, r: r'h: h', where h is the impetus corresponding to r, and h' to r'.

Hence,

h'
h'

r=r. and h' h

r

For r2h sin 2e; hence, r÷h, when e is given.

EXERCISE.-If a shell ranges 4000 feet, with an impetus of 1800, how far will it range with an impetus of 1980?

Ans. 4400.

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