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Integrating (3),

4

X 1200 cos 30°t + C3 and y 16t2+1200 sin 30°t + C1. To determine the constants we note that when t 0, x therefore C3 = 0, and C1 = 0.

0, y = 0, and

Hence

1612+1200 sin 30°·t.

(4)

X = 1200 cos 30°t and y

These equations give the position of the particle at any time. Solving the first of these equations for t and substituting in the second we find the equation of the path to be

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which is the equation of a parabola.

To find the range, let y = 0 in (5) and solve for x.

(5)

Hence

X

12002 cos2 30° tan 30°
16

38,970 feet

7.38 miles.

32t+1200 sin 30°.

1200 sin 30°
32

At the highest point = 0, hence from (3), 0

Therefore the highest point is reached when t 182 seconds. Substituting this value of t in (4), we obtain for the maximum height attained

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1. In the following exercises find the equation of the path in terms of x and y by eliminating t from the given equations, plot the path, find the magnitudes and directions of v and a for the given value of t, and at the point on the path corresponding to this value draw lines representing v and a in magnitude and direction.

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2. Find the equation of the path of a projectile fired at an angle of 20° to the horizontal with a muzzle velocity of 1500 feet per second; find the range of the projectile, the maximum height attained and plot the path.

3. Find the equation of the path of a projectile fired at an angle of 0° to the horizontal with an initial velocity of vo feet per second. Find the range and the maximum height attained. Find vo if the range is 24 miles for 0 = 45°.

4. A bomb is dropped from an aeroplane 8000 feet high moving horizontally at a velocity of 120 miles an hour. Determine how far the bomb

will fall from the point on the ground directly below the aeroplane at the instant it was dropped. Suggestion: If the point on the ground is chosen as the origin with the axes horizontal and vertical, the initial conditions are as given in the table

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5. A bomb is dropped from an aeroplane h feet high and moving with a velocity of v。 feet per second. If d represents the distance from a point on the ground directly below the aeroplane when the bomb was dropped to the point where it strikes the ground, find d as a function of h and v。. Plot the graph of d as a function of vo, h being constant, also of d as a function of h, vo being constant. In the first case what is the effect on d of doubling vo, and in the second case of doubling h?

6. A ball rolls off a roof inclined at 30° to the horizontal. If it starts 20 feet from the eaves, which are 30 feet from the ground, where will the ball strike the ground?

2t.

7. The path of a particle is given by the equations x = t2 and y Plot the path, find the acceleration when t 1, and the components of the acceleration along the tangent and normal lines to the path at the point in question.

8. Find the range of a rifle ball fired horizontally from a point 5 feet above the ground if the initial velocity is 2000 feet per second. How much is the deviation from the horizontal at 100 yards? 200 yards?

9. A particle moves in a circle whose center is at the origin of a rectangular system of axes. The angle (in radians) through which it turns from an initial position at rest on the x-axis is given by the equation D10 and the angular acceleration

0 t2 - t. Find the angular velocity w
=
απ Diw when t 3.

10. A wheel revolving at the rate of 120 revolutions per second is retarded uniformly so that in 3 seconds, w = 90 revolutions per second. How long before the wheel will come to rest, and where will a point P on the rim be if when t O the angle for P is 00?

109. Volume of a Right Prism. A prism is a solid bounded by two congruent polygons lying in parallel planes with their corresponding sides parallel and by the parallelograms determined by the pairs of corresponding sides of the polygons. The other sides of these parallelograms are called the lateral edges of the prism, the polygons are called the bases, and the parallelograms the lateral faces of the prism.

A right prism is one whose lateral edges are perpendicular to the planes of the bases.

A rectangular parallelopiped is a right prism whose base is a rectangle, e.g., an ordinary box.

A polyhedron is a solid bounded by portions of planes. The volume of a polyhedron is the ratio of the polyhedron to a second solid taken as the unit of volume. The unit of volume is usually a cube whose edge is the linear unit.

FIG. 182.

Theorem 1. The volume of a rectangular parallelopiped is the product of the area of its base and the length of its altitude.

If the three dimensions of the rectangular parallelopiped are integers, h, j, k, the number of unit cubes in it is easily counted.

In any horizontal layer in the figure, there are k rows of cubes, j in each row. Hence by the definition of the multiplication of integers, there are jk cubes in each layer. And since there are h layers, the parallelopiped contains hjk cubes.

Hence the ratio of the parallelopiped to the unit cube is hjk, so that the volume is

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be sawed into two congruent pieces, as in the figure. These pieces are triangular right prisms, whose bases are right triangles.

Theorem 2. The volume of a triangular right prism, whose base is a right triangle, is the product of the base and the altitude. If V is the volume of the triangular prism ABCA'B'C', then volume of AD'

1 V

} area ABDC XAA

area ABC × AA'

bh,

where b is the area of the base, and h is the altitude of the

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bases ADC and BDC are right triangles, as indicated in the

figure.

Then

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A

B

C

FIG. 185.

D

= ADCA'D'C' + BDCB'D'C'

= ADC × DD' + BDC × DD'
(ADC + BDC) × DD'

ABC X DD'

- bh,

where b is the area of the base and h the length of the altitude of the given prism.

Theorem 4. The volume of a right prism is the product of the area of the base and the length of its altitude.

Let AD' be any right prism. It may be divided into triangular right prisms by planes through one of the lateral edges, say AA', and each of the non-adjacent lateral edges. If V is the volume, b the area of the base, and h the altitude of the prism, we have

V = AD' = ABCB' + ACDC' + ADED' +
ABC × h + ACD × h + ADE × h +

(ABC + ACD + ADF +

= bh.

• )h

110. Volume of a Right Circular Cylinder. A cylindrical surface is a surface generated by a straight line which moves parallel to a given line so as to cut a given curve. The various positions of the moving line are called the elements of the surface.

A cylinder is a solid bounded by a cylindrical surface and two parallel planes. The sections of these planes cut out by the surface are called the bases, and the distance between them the altitude of the cylinder.

In a right cylinder the elements are perpendicular to the planes of the bases, and any element is equal to the altitude. If the bases are circles, the solid is called a right circular cylinder. Let r be the radius of the base and h the altitude of a right circular cylinder.

Construct a right prism whose bases are regular polygons of n sides inscribed in the bases of the cylinder, and a second

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prism whose bases are regular polygons of n sides circumscribed about the bases of the cylinder. Let V' and V" be the volumes of these prisms, b' and b" the areas of their bases. Their altitudes are equal to h. Then by Theorem 4, Section 109, V' b'h and V"

b'h.

Now let n increase indefinitely. Then by a theorem of plane geometry, b' and b" approach the area of the circle, Tr2, as a limit. Hence, by Theorem 2, page 266, V and V" approach the same limit r2h.

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