PROBLEM II. If a cannon ball be fired with a velocity of 1000 feet per second, up a smooth inclined plane, which rises 1 foot in 20: it is proposed to assign the length which it will ascend up the plane, before it stops and begins to return down again, and the time of its ascent. Heref as before. 4gf Then, by theor. 5, s= And, by theor. 7, t = =31088016 feet, or nearly 59 miles, the distance moved. 2gf 2 × 1672 × 20. 621" 147 10' 21' 147, the time of ascent. 193 1939 120000 = = 193 PROBLEM III. If a ball be projected up a smooth inclined plane, which rises 1 foot in 10, and ascend 100 feet before it stop: required the time of ascent, and the velocity of projection. FIRST, by theor. 6, v4gfs 4 x 16 X X 100 = 8/10 25 36408 feet per second, the velocity. And, by theor. 7, t =√/== 100 1672 × to 10 476 192107.88516 seconds, the time in motion. √10 = PROBLEM IV. If a ball be observed to ascend up a smooth inclined plane, 100 feet in 10 seconds, before it stop, to return back again: required the velocity of projection, and the angle of the plane's inclination, And, by theor. 8, ƒ == 16 × 100 193 is, the length of the plane is to its height, as 193 to 12. Therefore 193 12: 100: 62176 the height of the plane, or the sine of elevation to radius 100, which answers to 3° 34', the angle of elevation of the plane. PROBLEM PROBLEM V. By a mean of several experiments, I have found, that a cast iron ball, of 2 inches diameter, fired perpendi. ularly into the face or end of a block of elm wood, or in the direction of the fibres, with a velocity of 1500 feet per second, penetrated 13 inches deep into its substance. It is proposed then to determine the time of the penetration, and the resisting force of the wood, as compared to the force of gravity, supposing that force to be a constant quantity. 4gs 4 × 16 × = 32284. That is, the resisting force of the wood, is to the force of gravity, as 32284 to 1. But this number will be different, according to the diameter of the ball, and its density or specific gravity. For, v2 since ƒ is as by theor. 4, the density and size of the ball S remaining the same; if the density, or specific gravity, n, vary, and all the rest be constant, it is evident that ƒ will be as n; and therefore ƒ as when the size of the ball only is constant. But when only the diameter d varies, all the rest being constant, the force of the blow will vary as d3, or as the magnitude of the ball; and the resisting surface, or d3 or as d force of resistance, varies as d2; therefore f is as only when all the rest are constant. Consequently ƒ is as when they are all variable. dnv2 the strength or firmness of the substance penetrated, and is here supposed to be the same, for all balls and velocities, in the same substance, which is either accurately or nearly See page 264, &c, of my Tracts. so. Hence, taking the numbers in the problem, it is the value of ƒ for elm wood. Where the specific gravity of the the ball is taken 74, which is a little less than that of solid cast iron, as it ought, on account of the air bubble which is found in all cast balls. PROBLEM VI. To find how far a 24lb ball of cast iron will penetrate into a block of sound elm, when fired with a per second. velocity of 1600 feet HERE, because the substance is the same as in the last problem, both of the balls and wood, Nn, and F =ƒ; 5:55 × 16002 × 13 Dv's S Dv2 therefore = S dv2 41 inches nearly, the penetration required. PROBLEM VII. It was found by Mr. Robins (vol. i. p. 273, of his works), that an 18-pounder ball, fired with a velocity of 1200 feet per second, penetrated 34 inches into sound dry cak. It is required then to ascertain the comparative strength or firmness of oak and elm. THE diameter of an 18lb ball is 5'04 inches D. Then, by the numbers given in this problem for oak, and in prob. 5, for elm, we have From which it would seem, that elm timber resists moré than oak, in the ratio of about 8 to 5; which is not probable, ás oak is a much firmer and harder wood. But it is to be suspected that the great penetration in Mr. R.'s experiment was owing to the splitting of his timber in some degree. PROBLEM VIII. A 24-pounder ball being fired into a bank of firm earth, with a velocity of 1300 feet per second, penetrated 15 feet. It is required then to ascertain the comparative resistances of elm and earth. COMPARING the numbers here with those in prob. 5, it 1800 271 5.55 x 13002 × 13 = = 20 nearly 63 nearly. That is, elm timber resists about 63 times more than earth. 133 × 0.37 PROBLEM IX. To determine how far a leaden bullet, of of an inch diameter, will penetrate dry elm; supposing it fired with a velocity of 1700 feet per second, and that the lead does not change its figure by the stroke against the wood. HERED = 3, N = 11, n = 7÷. Then, by the numbers and theorem in prob. 5, it is s = = 93 inches nearly, the depth of penetration. But as Mr. Robins found this penetration, by experiment, to be only 5 inches; it follows, either that his timber must have resisted about twice as much; or else, which is much more probable, that the defect in his penetration arose from the change of figure in the leaden ball he used, from the blow against the wood. PROBLEM X. A one pound ball, projected with a velocity of 1500 feet per second, having been found to penetrate 13 inches deep into dry elm: It is required to ascertain the time of passing through every single inch of the 13, and the velocity lost at each of them; supposing the resistance of the wood constant or uniform. THE velocity v being 1500 feet, or 1500 x 12 = 18000 inches, and velocities and times being as the roots of the spaces, in constant retarding forces, as well as in accelerating 26 13 1 of a second, the whole time of passing through the 13 inches; therefore as |