(21) If the number of units of space, described by a body in the last second of its fall, be to the number of units in the final velocity, as 8 to 9, for how many seconds does the body fall?

The time of falling is 4 seconds.

(22) A body, acted on by a uniform force, has described 100 feet from rest in 2': in what time will it pass over the next 125 feet?

The required time is one second.

(23) Supposing gravity to act on a body during the 1st, 3rd, 5th, &c., seconds and not during the 2nd, 4th, &c.; shew that the space described from rest in 2t seconds is equal to

gt (2t+1).

(24) A stone, dropped into a well, is heard to strike the water after t seconds: find the depth of the surface of the water, the velocity of sound being assumed as known.

=

If u the velocity of sound, the required depth is equal to

(25) The velocity of a body increases from ten to sixteen feet per second, in passing over thirteen feet under the action of a constant force; find the numerical value of the force.

The numerical value is 6.

(26) Since (2fs)* is the velocity generated by an accelerating force f in a body moving through a space s, when there is no initial velocity, therefore, by the second law of motion, u + (2ƒs)* is the velocity, after motion through s, when there is an initial velocity u. Point out the fallacy in this argument.

(27) A body is projected vertically upwards with a velocity of 25 feet: determine its height and velocity at the end of two seconds.

At the end of 2 seconds the body is descending with a velocity of 39.4 feet, and its depth below the point of projection is 14.4 feet.

(28)

A body is projected vertically upwards with a velocity of a hundred feet: determine its altitude of ascent at the end of two seconds.

The required altitude is equal to 135.6 feet.

(29) A body is projected vertically upwards with a velocity of a hundred feet: determine the greatest height to which it will ascend, and the time of ascending to this height.

The required height is 155 28 feet and the required time of ascent is 3.1 seconds, approximately.

(30) A body is projected vertically upwards with a velocity of 3g feet: find its height and velocity at the end of four seconds.

The required height is 4g feet, and its velocity, which is downward, is one of g feet.

(31) A body is thrown vertically upwards with a velocity 3g: at what times will its height be 4g, and what will be its velocity at these times?

It will be at the height 4g, first, at the end of two seconds, and, again, at the end of four seconds: its velocity at both these instants is g, being upward at the end of two seconds and downward at the end of 4.

(32) A body is projected vertically upwards with a velocity which will carry it to a height of 2g feet: after how long a time will it be descending with a velocity g?

After an interval of 3 seconds.

(33) It is said that, on one of the asteroids, a man, who on the Earth could leap a height of 6 feet, could jump 60 feet high: compare the attraction of the asteroid on its surface with the force of terrestrial gravitation.

The Earth's attraction is ten times as great as that of the asteroid.

CHAPTER II.

PROJECTILES.

SECT. 1. Motion resolved horizontally and vertically.

(1) To prove that the time of describing any portion PQ of the parabolic path of a body, acted on by gravity, is proportional to the difference of the tangents of the angles which the tangents at P and Q make with the horizon.

Let V, V' be the velocities of the body at P, Q, respectively, a, a' being the inclinations of the tangents at these points to the horizon. Let u= the horizontal velocity of the body, and t the time of describing the arc PQ.

(2) Having given the velocities at two points of the path of a projectile, to find the difference of their altitudes above a horizontal plane.

Let u, v be the horizontal and vertical components of the velocity at any point A of the path: 'being the vertical component at a point P, h feet higher than A. The horizontal component will be the same at both points.

Similarly, V' being the velocity at a point P', h' feet higher

and therefore the difference of the altitudes of P, P', above a horizontal plane, is equal to

(3) If the focus of the path of a projectile be as much below the horizontal plane through the point of projection, as the highest point of the path is above it; to find the angle of projection.

If V be the velocity and a the angle of projection; the greatest altitude is equal to

V2 sin2 a
2g

Also the distance between the vertex of the path and the focus is equal to a quarter of the latus rectum, that is, to

(4) Particles are projected from the same point in the same direction, but with different velocities; to find the locus of the foci of their paths.

Let O, fig. (104), be the point of projection, A the vertex of the parabolic path, S its focus, OH half the horizontal range.

W. S.

18

Join OS.

Then

Let V the velocity and a = the angle of projection.

=

This result shews that the locus of S is a straight line through O making an angle with OH equal to π- 2α.

(5) If a body be projected in a direction inclined to the horizon, to prove that the time of moving between two points at the extremities of a focal chord of the parabolic path is proportional to the product of the velocities of the body at the two points.

Let P, P', fig. (105), be the ends of any focal chord PSP': draw PH, P'H', SK, vertically, to meet the directrix in H, H', K. Let PSK = 0, t = the time from P to P', u = the horizontal component of the body's velocity. Also, let V, V', denote the

velocities at P, P'. Then, since the velocity at any point of the path is that due to its depth below the directrix,

But

V2=2g. PH, V2=2g. P'H',

(V.V')2 = 4g2. PH.P'H'.

HP+SP cos 0= SK=H'P-SP' cos 0,

and therefore, since SP HP, and SP' = H'P',

=