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and time along pg = mga

PQ

Hence, the whole time through MPQ is

PQ ve +

- + 2a - x) 9

2gx and that through Mpq is

PQ 20

(2+ w 20-x) g

29 2a - x which being identical, are equal.

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2s

331. Let (s) be the space required ; then the whole time of descent (t) is expressed by

✓ Now since S =

9.To generally the space described in the 4th second is

. (49 = 30) = 11/25 and that described in the last second but 4, is 2. {(t – 4)2 – (t – 5)} =

g (2t - 9)

(

2 Hence, by the equation,

: 2t 9 :: 1 : 3 it = 15

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332. Let t be the time required. Then by the expression for the circumstances of motion on inclined planes,

gta sin. 6

2
we have by the question

3
8 = 30 feet, = 30°, g = of the force of gravity =

S =

х

& 을 (32 feet) = 24 feet nearly

8

60 X 2

= 4,9740932
24-

8
it = 2.23 seconds nearly.

333. Let Aa, Bb (Fig. 8) be the perpendiculars to the horizon bt. Then, joining AB let it be produced to meet ba in T, and take TP a mean proportional between TA, TB, and describe the circle APB passing through A, P, B, and join AP, PB, P is the point required.

For since TP = TA X TB the circle touches the horizon in P.

Hence time down BP = time AP. The point B will fall within or without ab according as the arc

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334. Let AB = a, AC = b, and the distance fallen by the latter body before it is overtaken = x; then

6 a + x is the distance fallen by the former in the same time t with the velocity (v) acquired at B. Hence 6 – a + x = tv + to

= tv2ga + .

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335. If o be the inclination of the plane, the moving force with which P descends is

P - Q sin. O
:: the accelerating force () is

P-Q sin.

P + Q Now V = got = g.

P-Q sin. O

x t the velocity of P acquired

P+Q in the time t, the inclination of the plane being any whatever. In the problem, 6 = 30°.

2P - Q 2 (P +Q)

:: V=g

336. Let t be the time in which the bodies meet, a the distance between the points of projection, and from the upper of these let x be the distance of the point of concourse; then

x = tv +

9 2

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d'e 28 = ga = (W ga )' = the square of the velocity

Ve acquired down

since s = .2

29

a

337. Let a be the inclination of the given plane to the horizon, 8 of that sought; then the moving force is P. sin, a

W sin. 0
and :: the accelerating force () is

P sin. a - W sin. A
P + W

I
Now | OC

Х sin. O

P sin. a – - W sin. O :: P sin. a sin. 0 W sin.” 0 = max. and putting the differential = 0, we get

P sin. sin. 0 =

2W giving the inclination required.

= min.

a

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338. Here the moving force is evidently 1 oz. and ;. the

1
accelerating force is p = by avoirdupoise weight.

33
Hence $ =
9

9

ť 66

3302

290 24
But s by the question = 12 feet.
12 X 66

33

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and se

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340. Let a be their distance at first, x that after the lapse of the time t; then since the spaces described in that time are tu +

9

f', and tv 9.t

2

2

we have
x = t.(v + 0)

(1) Now when the bodies meet x = 0, and . t =

....

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if for t in equation (1) we put

a

otor

, there results, after reduction

a

a the distance required.

34).

The moving force being

H
P - W.

P

by the question, we have L 6

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342. Here the moving force is 1 oz. and, by the question, we have, therefore,

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