value must be assigned to the accelerative effect of gravity? The method of obtaining this value will be readily understood by considering the following Examples : Ex. 661. What velocity would be acquired by a body that fell freely for one minute? Ans. 1920 ft. per sec. Ex. 662. If a body moves with a velocity of 1920 ft. per second, what Ans. 38,400. is its velocity estimated in yards per minute? [Now, it must be remembered that at the end of each second a falling body acquires an additional velocity of 32 feet per second; it appears from the last two Examples that the same body would acquire in each minute an additional velocity of 38,400 yards per minute; but the former number represents the accelerative effect of gravity in feet and seconds, and therefore the latter represents the accelerative effect of gravity in yards and minutes.] Ex. 663. If the unit of space were a fathom, and the unit of time 15 sec. what would be the numerical value of g? Ans. 1200. Ex. 664. Given that the accelerative effect of gravity at the distance of the moon equals 11 in feet and seconds, find its value in miles and hours. Ans. 21.91. Ex. 665. An accelerative of a certain force has a numerical value f when certain units are employed: show that its value will be m2 f when the new n unit of time contains m, and the new unit of space n, of the old units respectively. FIG. 157. A B 127. Composition of Velocities. - Suppose a body to be at any instant at the point a, and suppose it to be moving with such a velocity as would in a certain time carry it to B along the line AB; suppose that at that instant another velocity were communicated to it such as would in the same C D time carry the body along the line AC to c, if it had moved with that velocity only; complete the parallelogram ABCD and join AD, then at the end of the given time the body will arrive at D, having moved along the line AD. That this is so appears at once from the well-known fact, that when a ship is in a state of steady motion, a man can walk across her deck with precisely the same facility as if she were at rest; thus if he were to walk across the deck when the ship is at rest he would go from A toc; but if we suppose the ship to have such a velocity as will in the same time carry the point a to B, he will come to the point D; and if the velocities have been uniform he will have moved along the line AD. Now, let v and u be the two velocities, then AB:AC:: u: v and if v is the velocity compounded of them AD:AB:: ν: υ So that if AB and AC represent the given velocities in magnitude and direction, A D will represent the velocity compounded of them in magnitude and direction. Hence the rule for the composition of velocities is the same, mutatis mutandis, as that for the composition of forces. If the velocities vary from instant to instant, the rule will give the magnitude and direction of the component velocity at any instant; this is the case which commonly happens, for example, when a body is thrown in any direction transverse to the action of gravity; the method of determining the motion of the body may be described in general terms as follows: Conceive the time to be divided into a great number of intervals, and suppose the velocity that is actually communicated by gravity during each interval to be communicated at once,* then, by the composition of velocities, we can determine the motion during each interval, and therefore during the whole time; the actual motion is the limit to which the motion, thus determined, approaches when the number of intervals is increased. * It is immaterial whether we conceive it to be communicated at the beginning or at the end of the interval. Proposition 25. A body is thrown in vacuo with a given velocity (v) in a direction making any angle with the horizon, to determine its position at the end of any given time (t). Let the body be projected along the line AN; take AN equal to vt, and divide t into n parts, each equal ; then if AN is divided into the same number of equal parts in During the first interval the body will move over the space AN1; draw N1n, vertical and equal to gTxT; complete the parallelogram n12; then since the sides Nin, and NIN2 are proportional to the velocities gr and v, the body will, during the next interval, move along the line N1Q, and at the end of the interval will arrive at a point q vertically under N2; the actual velocity with which the body has moved being, of course, equal to N1Q-T. At the point q we have to compound this velocity with gr; to do this we must produce N1Q to r, making or equal to NQ; take on, equal to gтхт, and complete the parallelogram, then the sides of this figure are proportional to the component velocities, and therefore the diagonal is in the same proportion to the velocity compounded of them; at the end of the third interval therefore the body will be at R vertically under N3; the same construction will apply to any number of intervals, and the required point p will be vertically under N. To determine NP; produce NQ to m1, QR TO M2, RS to m3, &c., then will NP equal the limit of the sum of Nm1, m1m2, m2m3, &c.; but by similar triangles Nm, is the same multiple of N2Q that NN is of N1N2, therefore Nm, equals (n-1) дт2, similarly m1m, equals (п-2) дт2, т2m3 equals (n-3) gr2, &c., and therefore their sum equals (n-1) дт2+(n-2) дт2+(n-3) дт3+...+2дт2+дт2 =gr2{(n-1)+(n-2)+..+2+1} =92 • 2 Now, however great the number of intervals, Q, R, S, &c., will remain vertically under N1, N2, N3, &c., so that in the limit p will remain vertically under N. Also the limit of gt2(1-) is gt; so that the true position of the body will be found by measuring downward from N a distance equal to gt2.* Ex. 666. A point moves along a smooth horizontal plane with a velocity of 3 ft. per second; at the end of 2 seconds a velocity of 8 ft. per second is impressed on it in a direction at right angles to its motion; after how long will its distance from the starting-point be 20 ft. ? Ans. 4 sec. Ex. 667. A body is projected in vacuo with a given velocity in a given direction; determine its range on a horizontal plane passing through the point of projection and the time of flight. * This result is, of course, true for any constant force (f) acting along parallel lines. Hence, if a body has at a certain point a given velocity (v) along a given line an, and is acted by such a force, its position at the end of t seconds is given by the construction-take an equal to vt, drawn NP parallel to the direction of f, take NP=2ft; P is the required position of the body. Ex. 668. A body is projected with a velocity of 100 ft. per second in a direction making an angle of 37° with the horizon: determine the time of flight and range on a horizontal plane. Ans. 3.76 sec. and 300.4 ft. Ex. 669. If a body is thrown with a given velocity the horizontal range is greatest when it is projected at an angle of 45°; and for angles of projection one as much less as the other is greater than 45° the horizontal ranges are the same. Ex. 670.-Show that the least velocity with which a body can be projected to have a horizontal range R is 4√2R feet per second. Ex. 671. Determine the angle of elevation and velocity of projection that will enable a body to strike the ground after 10 seconds at a distance of 5000 ft. from the point of projection. Ans. (1) 17° 45'. (2) 525 ft. per sec. Ex. 672. A body is projected with a velocity v in a direction making an angle a with the horizon; if R is its range on a plane passing through the point of projection and inclined at an angle 0 to the horizon, and the time of flight, determine R and r. FIG. 160. N A M B Let a be the point of projection; draw Ama horizontal line through A, AB the inclined plane, AN the direction of projection; let the projectile strike the plane at B, then we have |