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Also, since (6 being > a)

yo'l – y: y' - y :: B: A + B
and I" - *::-.::B: A + B
we have
y" (A + B) = y'B + yA2

" (A + B) = x'B + XAS And eliminating x, t', y', y' from the above five equations (1), (2), (3), we finally obtain

aA sin. 0 + 6B sin. y" =

Q

g(A + B) cửa 0 + bB cos.o 2(aA cos.0+ bB cos.c) the equation to a parabola, whose greatest ordinate, corresponding abscissa, and parameter, are easily found to be (aA sin. 0 + bB sin. p)", (aA sin. 0 + bB sin. ) 29(A + B)

1 x (aA cos. 0 + bB cos. 9) X

and

g(A + B) 2(aA cos. 0 + bB cos.o)

respectively. 9(A + B): No.(374) is but a particular case of this.

In the problem, A = 2, a= 20, 0 = 60°; B = 3,6 = 25, and Q = 30°, whence it is easy to find the required quantities.

390. Since the horizontal velocities of both sound and the projectile are constant, and describe (by supposition) the same space in the same time, these velocities must be equal (s = tv). Hence, a being that of sound, we have, by the question, and by '370)

gt' y = 0 = x tan. O

2a'cos. 28

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where, by the question, y = 0,6 = 15, x= 100 feet,
Hence, substituting, &c., we get
22.

= 100 feet, height required.
2g

X t?

393. Generally, we have

8

H
Stv

2 L since the space described with the velocity v continued uniform would be tv, and that through which the body would descend

H down the plane t". But by the question, v = 30 feet,

2

L

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.. S = 360 12g = 26 feet nearly, or the body will be lower down the plane by 26 feet than the point of projection.

Hence the velocity (u) of the body after 12" will by 30 + that acquired by falling through 26 feet of the plane.

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396. By (370) and the question, we have
0 = y = 150 tan. 30°

g(150)

2vo cos, 300'
which gives
1

3
10

9

tor

397. By (370) and the question, y = x tan. 60°

gx?

2(100) cos.' 60° = x N 3

gx?

5000 Hence, if 6 be the distance of the foot of the inclined plane from the point of projection, (since the equation to the section of that plane made by the plane of motion is y' = (x' - b) tan. 30°

', x

= (1 - ) ).

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Again, the greatest altitude above the horizon is (370)

3750 feet, whence it is easy

va

' 2g

8g to find that above the plane.

9

398. By (389) the greatest altitude of the centre of gravity of A, B, projected with the velocities a, b, is generally,

(aA sin. 0 + bB sin. c)'

29 (A + B)
Here 0 = Q = 90°; :: the greatest altitude required is

(aA + bB)
29(A + B)

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sin. %

29 Also

sin. 28

29 Consequently, if

y' = at' + 6 be the equation, referred to the same origin, of the section of the inclined plane, made by the plane of motion, we have

y - y = (sin. 28 - a sin. 20) - 6

2g for the greatest altitude required.

400. Making the intersection of the horizon with the bank, the line of abscissæ (x) (originating in the point of projection), and taking the ordinates (y) in the plane of the bank, it is evident that the motion parallel to (x) will be uniform, and that parallel to (y) the same as down the common inclined plane, whose inclination («) is the same as that of the bank. Hence, (u) being the velocity of projection, and 0 the which its direction makes with the line of intersection, we have,

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g sin.c x = x tan. A

2v? cos. 20 the equation to a parabola.

The greatest ordinate, corresponding abscissa, and parameter, are easily found to be

sin. 20

sin. 20 29

29

sin. a

sin. a

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sin.

a

I The place where the body meets the horizon is determined by putting y = 0, which gives

v2 sin. 20

y =

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401. Let AB = BC (Fig. 89.) represent the equal times of descent and ascent; Aa, Cc the initial velocities, and Bb, BD the ultimate velocities; then the spaces described (Wood, p. 139,) are denoted by

ABba, BCcD. But, by the question, the spaces described are equal, and abb = DEC (being the spaces due to gravity). Hence

ABDa = BCED; and since

AB = BC, .. Aa = BD = CE. :: Cc – Aa = Ec = velocity acquired by falling vertically in the time AB

402. Let a, b, be the altitudes fallen through by A, B, to acquire the velocities u, v; then

u= 2ga,

2ga, v= 2gb Hence, being the horizontal distance moved, we have

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