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EXAMPLES. XXX,

1. Prove that the interval occupied by a particle in sliding down an inclined plane of height h ft. and inclination a is

( 12h)

I

sin a

2. A train is on a smooth incline of 1 in 100; find how long it will take to go a mile from rest. [N.B. by an incline of 1 in 100 is meant that in the figure of page 91, the incline is such that BC=1 and AB=100.]

3. Find the velocity acquired at the end of the mile by the train in Example 2. If another train start to go up a smooth incline of 1 in 50 with that velocity, shew that it will go half a mile up the plane.

4. A particle slides down a smooth straight tube of length 1 inclined at an angle a to the horizon, and then falls freely under the action of gravity; if the lower end of the tube is h feet above a horizontal plane, find when and where the particle will strike the plane.

5. A particle slides down a smooth straight tube, whose upper point is at a height h ft. from a horizontal plane, and then falls freely under the action of gravity; prove that it will in all cases reach the horizontal plane with V/2gh) velos.

6. A particle is sliding down a smooth tube inclined at an angle to the horizon, and it takes twice as long to descend any vertical distance as a particle falling freely; find the inclination of the tube.

7. Find the direction in which a smooth tube must be drawn, so that one end being at a fixed point and the other end on a fixed straight line, a particle sliding down it may reach the straight line in the shortest possible time.

NOTE. The above problem is often stated thus. Find the line of quickest descent from a given point to a given straight line.

8. Prove that the time of sliding down any chord, supposed smooth, of a vertical circle drawn to the lowest point, is constant.

9. Find the line of quickest descent from a given straight line and a given point.

*SECTION III.

ILLUSTRATIONS.

CHAPTER IX.

PROJECTILES.

117. A particle which, having been projected from any point in any direction, is then supposed to move under the action of its own weight only, is called a projectile.

The resistance of the air is here neglected. This however has a very considerable effect on the motion of any mass actually projected, and renders this solution of the problem of little practical use.

118. When a projectile is projected vertically upwards its path is a vertical straight line.

This case has already been considered [Art. 46).

119. When a particle is projected in a direction which is not vertical, its path is a curved line.

This curved line is in one plane, which contains the vertical and the original direction of projection. For there is no force tending to take it out of this plane.

We shall denote the velocity of projection by u velos, and shall suppose that its direction makes an angle a (radians) with the horizon.

The resolved parts of the velocity of projection

I 20.

I 21.

are

(i) u cos a (velos) in the horizontal direction,
(ii) u sin a (velos) in the vertical projection.

L. D.

7

The horizontal velocity of a projectile is

122. PROP. constant.

By the horizontal velocity is meant the resolved part in a horizontal direction of the velocity at any instant.

The acceleration of the particle is always vertically downwards; so that after any interval, the additional velocity of the particle is also vertically downwards, and consequently has no resolved part in the horizontal direction.

The horizontal velocity is therefore unaltered. Q. E. D.

123. PROP. The vertical velocity of a projectile after t seconds is

(u sin a- gt) velos. This is the expression of the fact that initially the particle has a vertical velocity u sin a velos, and that it has the vertical acceleration - g celos.

124. PROP. The velocity of a projectile after t seconds is

_{(u cos a) + (u sin a - gt)*} velos, and if o be the angle which its direction makes with the horizon

u sin a-gt tan A

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For, the velocity of the projectile at any instant is the resultant of the horizontal and vertical velocities at that instant (See Art. 95).

125. PROP. To find when and where a projectile is at its greatest vertical height.

The vertical velocity is (u sin a - gt) velos.

When this is positive the particle is going upwards, and when this is negative it is going downwards, and when this is zero it has reached its greatest height; that is, when

u sin a

The vertical distance is

(u sin at – įgt) feet,

u sin a in this putting for t, we find for the greatest vertical

8

2 sin a distance

feet.

28 The horizontal distance at the same instant is found by

u sin a
putting fon in u cos at; it is
8

u* cos a
a sin a

feet.
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1. A particle is projected under the action of gravity with 25 velos making an angle whose tangent is with the horizon; find when it is moving with 20 velos.

2. A particle is projected with 416 velos making an angle with the horizon whose sine is 1* ; shew that its least velocity is 160 velos, and that when it has that velocity it is moving horizontally.

3. Find when the projectile in Example 2 has a velocity of 384 velos ; find also its direction of motion at that instant.

4. Shew that the total velocity of a projectile projected with given initial velocity at a given angle is least when it is moving horizontally.

5. A projectile is projected with u velos making an angle a with the horizon; prove that it will be moving in a direction at right angles to the direction of projection after cosec a seconds.

u

6. Two particles are projected simultaneously in the same vertical plane with velocities u velos and v velos at angles a and B to the horizon respectively; shew that their velocities will be parallel after

uv sin (a - b)

seconds. g (v cos ß- u cos a) 7. A particle projected with u velos at an angle a, is moving in a direction making an angle ß with the horizontal plane after

(u sin a u cos a tan ß) seconds.

1 26. PROP. The path of a projectile is a Parabola.

Let u velos be the initial velocity making an angle a with the horizon.

Let P be the point of projection ; Q the position of the projectile after t seconds. Draw PP' horizontal; draw PV vertically downwards; draw PH making an angle a with PP'; draw QH vertical and QV parallel to HP.

Then (Art. 91] PH= ut ft.; HQ = £gt* ft.

= u*

Therefore QP° =PH° = oto

2HQ 2u

PV 8 8

2u = 4SP. PV; where 4SP

ft.

8 But in the Parabola, QV = 4SP.PV; where SP is the distance of the focus from P.

Therefore, if we draw a parabola, touching PH at P, whose axis is vertical and vertex upwards, such that

2u 4SP

ft.,

then Q will always be on this parabola. Q. E. D.

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