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:: x = ab .. the time of B's uniform motion is 2N ab

the time of A's descent. 2gb

9

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v

403. Let o be the inclination of the line to the horizon, then the time down it is

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- 28.02

v2

= x3 But by the question,

X = 300 x cos. 30° = 150 3 and

y

– 300 X sin. 30o = 150 whence v = 15 2g. Hence the greatest altitude above the plane is easily found, by the rules of max. and min., to be 05 feet; and the time of flight

150 N 3

= 10

V cos. 60

405. Let a be the inclination of the given plane to the horizon, a the vertical distance between the given point and the plane; also suppose the co-ordinates (x, y) of the required locus

to originate in the intersection of this vertical with the plane, s * being measured I and y parallel to the intersection of the plane with the horizon; then (t', y) being the horizontal and vertical co-ordinates of the path, described by the body projected parallel to the horizon, we easily obtain

y' = a - 2 sin. a

2'2 = y + xocos. « And by 370,

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gx 20%

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C

Hence
2ava 2v2 sin, a

20cos. Sa
9

g the equation to an ellipse, whose transverse and conjugate axes are 2v

2v 2ag + va tan.'a, and 2ag +vo tan.'« respectively. g cos. a

g

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the equation to a circle, whose radius is tv, and the co-ordinates of whose centre are

O and tv

gt?

Let p be the parameter of the given parabola, whose

408. equation is

= py

1

*, and y being the horizontal and vertical co-ordinates, originating in the vertex; then, supposing (a,b), (a', ') the co-ordinates of the given points of projection, the 2 (a, a) of projection, or the inclinations of the tangents to the parabola, at those points are dy

= tan. O) determined by

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And :: v' = 2gb (1 + eos. ) = 2gb (1 +

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d'? = 2gb (1 + cos, 'a') = 2gb' (1 +

4a'! Also, at the point of concourse,

a - x = tv cos. a, a + x = tv' cos. cé
b - y = tu sin. « - t", B' y = tv'sin, a'-

=

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409. Let m denote the degree of elasticity; then, 0, 0, X, being the 2 of reflection made with the radii, we have, (Wood, p. 131.)

tan. 0 = m tan.

tan.Q = m tan. x
and 0 + 0 + x = (radius of O = 1)

2

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410.

Let o be the inclination of the plane AB to the horizon, a, b the horizontal and vertical co-ordinates of the point A referred to the point B; then, v denoting the velocity at B, or that with which the body is reflected by the vertical plane in a direction inclined to the horizon by the 20, the trajectory described, referred by (x, y) to B, will be (370).

g.ro y = x tan. A

2v2 cos. 20 But, by the question, the body strikes the plane again in A;:. - b = a tan. 0

ga?
202 cos. 20

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But b = a tan. 0, a = N a? +63 x cos. 0.

la? +62

ag
x
b

2 sin.20

or

411. Let V, V, V .....denote the velocities of projection after each rebound, 0, 04, 0.......the elevations of projection. t, ti, to .......the times of flight. 7, 71, 7 ....... the horizontal ranges. h, hl, he......the greatest altitudes, a, Q ,, ....... the areas of the parabolas. VOL. II.

T

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cos.bg

Then, since the velocity of incidence on the horizontal plane is = that of the reflection immediately preceding, and (Wood, p. 131.) cos. O

cos. , V, Ug =

&c. = &c. cos. and tan. 0, = n tan. 0, tan. 0, = n tan. 01, &c. = &c.

:: (370)

sin. 20, r, = 0, sin. 20, &c., become by substitu2g

29 tion, and reduction.

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