(9) A heavy body has fallen from A to B, when another body is let fall from C, a point lower than B in the same vertical line: how far will the latter body fall before it is overtaken by the former?

Let AB-a, BC=b. Then, at the instant when the latter body is let go, the velocity of the former body is (2ga)*. Since gravity cannot affect the relative velocity of the two bodies, (2ga) is therefore always their relative velocity: hence, b being their initial distance, and t the time between the commencement of the second body's motion and the collision of the two bodies,

Hence the space fallen through by the latter body is equal to

The solution may be effected also in the following manner. Let s be the required fall of the latter body: then

s = gt2,

and, bs being the space described by the former in the time t, b+s = (2ga)1t +1⁄2gť3,

(10) A body, moving in a straight line, has velocities v1, v2, after intervals of one second each: in what case is the force uniform, and what is then its measure?

...

The velocities V1, V2, V3 must form an arithmetical progression, the measure of the force being the common difference.

(11) If a heavy body fall from rest through 144 feet, determine the time of the motion.

The required time is approximately equal to 3 seconds.

(12) A body, dropped from the top of a tower, the height of which is 60 feet, reaches the bottom of a well, at the foot of the tower, in 3": find the depth of the well.

The depth of the well is 84.9 feet.

(13) Determine the height through which a body will fall in two seconds and a half, and the velocity acquired.

The required height and velocity are respectively, in feet, 100 625 and 80.5.

(14) A, B, C, D are points in a vertical line, the lengths AB, BC, CD being equal: if a body falls from A, prove that the times of describing AB, BC, CD are respectively as

1: √√2-1 : √√3 – √2.

(15) A body describes in successive intervals, of 4 seconds each, the spaces 24 and 64 feet, in the same straight line: determine the accelerating force and the velocity at the beginning of the first interval.

The measure of the accelerating force is 2 feet 6 inches, and the measure of the required velocity is one foot.

(16) A particle moves over 7 feet, in the first second of the time during which it is observed, and over 11 and 17 feet in the third and sixth seconds respectively: prove that these facts are consistent with the supposition of its being subject to the action of a uniform force.

(17) A body, starting with a given velocity, moves for a given time under the action of a uniform force in the direction of its motion: shew that an equal space would be described by a body moving uniformly during the given time with a velocity equal to half the sum of the initial and final velocities of the first body.

(18) A falling body is observed, at one portion of its path, to pass through n feet in r seconds: find the number of feet described in the next r seconds.

The required number of feet is equal to n + gr2.

(19) A falling body has a velocity u, at first, and a velocity v, at the end of t seconds: shew that the space described is equal to

It (u + v).

(20) If s, ms, be the spaces described by a body in times. t, nt, respectively, determine the magnitude of the accelerating force and the velocity of projection.

(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 (2ƒs)* 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 + (2fs)* 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: v' being the vertical component at a point P, h feet higher than A. The horizontal component will be the same at both points.