tional to the force, as all bodies are with respect to their gravity, then the space is as the square of the time; for the ft2 if F, the accelerating sxt; if F: expression 8x, becomes sœť2; force, then s∞ Fi; and so on for other quantities. MISCELLANEOUS EXAMPLES. Ex. 1. What is the difference between the depths of two wells, into each of which should a stone be dropped at the same instant, the one will strike the bottom at the end of 6 and the other at the end of 10 seconds? Answ. 1029 feet. Ex. 2. In what time will a musket ball dropped from the top of Salisbury steeple, said to be 400 feet high, reach the ground? Answ. 5" nearly. Ex. 3, A heavy body was observed to fall 100 feet in the last second of time; from what height did it fall, and how long was it in motion? Put the time sought=r, and 16 feet=g; then, by art. 275, and the conditions of the question, we have gr2-gx x-1=100; that is, 2gx-g=100, and a 100+g. 2g 100+163 321 seconds, for the time required. Hence, the height from whence it fell is easily found to be 209127 feet. Ex. 4. A stone being let fall into a well, it was observed that, after being dropped, it was 10 seconds before the sound of the fall at the bottom reached the ear; what is the depth of the well? Put for the depth of the well, g=16 and b=1142 feet, the velocity of sound; then, by art. 274. ✔g:√x:: 1": g the time of the stone's falling to the bottom of the well; and since sound moves uniformly, :*:: 1": the time in which the sound ascends from the bottom; but the sum of these times is equal to the whole time: that is ✔=10,” by the question; from whence #1270 feet, nearly. Ex. 5. A stone descending from the top of a tower was observed to fall half the way in the last second; what is the height of the tower, and how long was the stone in falling to the ground? Answ. the tower's height is 187,48 feet, and the time 3.414". Ex. 6. In what time will a musket ball dropped from the top of Boston steeple, said to be 300 feet high, reach the ground? Answ. 4,32 seconds nearly. Ex. 7. A drop of rain, in its descent towards the earth, was observed to fall through a space of 400 feet in the last two seconds; from what height did it fall? Answ. 837,83 feet. On the motion of bodies on inclined planes, and curve surfaces; and the doctrine of the Simple Pendulum. PROPOSITION LXXII. 297. The force which accelerates or retards a body when moving upon an inclined plane, is to the force of gravity, as the height of the plane to its length. For, in fig. 122, let ac represent the absolute weight of the body; then ab perpendicular to AC, is the pressure against the plane, and is counteracted by the plane's reaction; but bc is the force which urges the body down the plané: therefore, the accelerating or retarding force, is to the force of gravity, as bc: ac. But the triangles abc, ABC, are similar, the sides of the former being respectively perpendicular to the latter; conse quently, be: ac :: BC: AC. 298. Cor. 1. Put the height of the plane BC=H, and it's length ACL; then, the accelerating or retarding force H F, on the plane, is F=, the force of gravity being repre sented by unity: for F: 1 (force of gravity) :: H: L. 299. Cor. 2. By trigonometry, H: L:: sine.<CAB H rad.=1; therefore, sine<CAB, or the force on the plane varies as the sine of CAB. H 300. Cor. 3. Because the force F, or, on the plane CA is in a given ratio to the force of gravity, it is a uniform force: H therefore, substituting for F, in the theorems 5, 6, 7, and 8 And by these theorems we can determine the motions of bodies on inclined planes, in the same manner as when the bodies move in free space, EXAMPLES. Ex. 1. If the length of an inclined plane be twice its height; to find the space through which a body will descend on this plane in 3 seconds. Н H Here [=), and s==×gt2=1×16×9′′=72} feet, is the space required. Er. 2. If the length and height of the plane be as in the preceding example, and a body descend from rest till it has acquired a velocity of 30 feet per second: the space described, v2L 900 × 2 or & = 4gH 4×16×1 =27,97 feet nearly. Er. 3. If a body fall from rest 12 feet on the above plane, to find the velocity acquired. Here H ✔4gs)=√(} × 4 × 16×12)=19,6 feet, L nearly per second. Ex. 4. Required the velocity a body would acquire by falling from a state of rest down this plane in 3". ' H Here v=x2gt×2×16×3=481 feet per second. Ex. 5. In what time would a body descend 12 feet from rest down the same plane? In this case t 8L gH 12×2 16×1 1,22 seconds nearly. Ex. 6. How long must a body descend from rest on the same plane to acquire a velocity of 481 feet per second? Ex. 7. A body descended 12 feet from rest along an inclined plane in 1,4922 seconds; what was the height of the plane, its whole length being 30 feet? Here H S 12 Lgt 16x(1,22) ̈16 ̧1⁄2 ×(1,22)2==1; therefore, the height of the plane is half its length=15 feet. PROPOSITION LXXIII. 301. If two bodies descend from the highest point A of an inclined plane ACB at the same instant, one of them will fall through the perpendicular altitude AB, while the other descends along the plane to D; where BD is perpendicular to AC. (fig. 123.) For complete the parallelogram BAED; then, it is evident that the body on the plane at A is acted upon by two forces, viz. gravity in the direction AB, and the reaction of the plane in the direction AE, perpendicular to AC; therefore, their joint actions will cause the body to describe the diagonal AD, in the same time that gravity would have caused it to describe the space AB. 302. Cor. 1. On AB (fig. 124.) describe the semicircle ADFB, where AB is perpendicular to the horizontal line BE; and from the highest point A, draw AE, AC, meeting the |