SOLUTIONS TO MATHEMATICAL PROBLEMS

Otago Witness, Issue 1489, 29 May 1880, Page 23

SOLUTIONS TO MATHEMATICAL PROBLEMS

Published May 7th. 265. By R. T., Nugget Point Lighthouse .•— 24 x 12 x 24 x ',18 = 124416 cubic in. 1728 • 124416 : : (55 x 16) : 63360 = the weight of pile in ounces. Then 1000 : 172S : : 63360 : 10948G 08= cubic inches of timber immersed. Therefore 109486 08 ~ (24 x 12 x 21) = 15 84 inches, the depth it will siuk on its broad side, and 109186 08-r (24 xl 2 x 18)=21;12 inches, the depth it will sink on its narrow side. 266. By E T., Nugget Point Lighthouse :— (6x6x6x9)v64==3o|lhs, the weight of iron ball. (6x6x6x2|-7-9=4Blbs= weight of lead ball. 48 - 30 a ! = 172 1bs difference.

ANSWERS. E. A. H., Palmersfcon, writes:— 267, husband's age, 45, wife's ago, .15 ; 261, mineral, ,

B£d, vegetable, lHd. In ref erenoe to 257 :— W. A., Grovefcown, Marlborough, in your issue of May Bth, senda solution by algebra of problem 257 which appeared in your issue of March 27th. The problem was as follows : The difference of two numbers is 9, and the quotient of the greater by the less is 7. What are the two numbers? I send you another solution which seems to me to satisfy the problem better and to be much simpler. Solution : — Let % se+9 =lesser number #+ 9 =greater =7 #+9=

x 7x. 8-7a}=-9 -6»=-9 6»=9 tf=l'6lesßer number. 15+9=10-5 greater number. Proof 10 ; 5-l*5=9, as per.problem; 10-5-rl's.=7, as per problem. Kiwi, Dnntroon, writeg :— R. T.s problem, 259. —A man falling from the mast-head, with the ship going 10 or 11 knots an hour, would naturally fall " according to the lawa of gravitation." Yet, undoubtedly, another influence would be at work to prevent his falling in a vertioal line, viz, the horizontal velocity of the Bhip, which velocity would be impressed on the falling man, and cause him to fall, not behind the mast, but at the foot of it, supposing the mast to be perpendicular. This _ has been verified at soa. A similar kind of thing may be seen on land. From a Chmtchurch express, I once aimed an apple straight for a youngster some 25 yds abreast of she carriage platform on which I was standing, and was amused to notice the curious figure the apple described. Instead of reaching the boy, it fell at least 20 yds to the side of him in the direction I was travelling ; thus proving that two forces were at work in the apple's motion; one, the attraction of gravitation ; the other, the horizontal velocity of the railway train. I should be aorry to make a practical experiment of falling fron a mast-head, but I am convinced I should fall in the way indicated by G. E O. O. D , Wyndbam, writes :— "With reference to the problem 259, givou by R. T., Nugget Lighthouse, in your issue of April 17th, G. E. 0., Christchurch, is perfectly correct in your issue of May Ist; ami J. Whitty, Southland, in your last week's, is entirely wrong in. his supposition that as soon as the man let go the mast he would lose his horizontal force. An equestrian in a circus has simply to spring upwards to paas through a hoop, If he sprang forward he would fall in front of the horse, Aud again, a rifle bullet fired vertically from a train going 60 miles per hour would (allowing a dead calm and the rifle set dead " plumb ") fall in the muzzle again. lam sorry I cannot give the curve the bullet would describe, but some of your readers may kindly say whether it would be a parabola or not, but the face of the bullet again falling in. the bore 13 a well-known one in dynamics. Supposing, as R. T. writes in your last week's issue, the man did loose his horizontal motion on letting go the mast, he would require to give the man's weight before this extravagant problem could be solved. D. C , Sandymount, writes : — Problem number 264, Fitz Barry, Kuri Bttßh, has not found the true answer, the propounders being too little by more than one-fifth of the true answer, inasmuch as the log in question was in reality the fruatrom of a cone. W. A., Grovetown, replies :—267,: — 267, B£d and ll|d ; 268, 15 and 45 respectively.