Now the greatest velocity that the man can acquire in his descent is that which takes place when the retarding force = the accelerating force. Hence the velocity is whence by equation (2) the time of acquiring this velocity is oo. In this solution the resistance of the air upon the man's body is not considered; so that it holds good but for the descent of the solid semicircle or cylinder whose weight is w. 736. The resistance a area x density × (vel.) .. R: R': A x D x V2: 4A × 3D × 4V2 :: 1 : 48. 737. Let r be the radius of the sphere, y that of the smaller end of the required segment; then if R denote the resistance on the circle ar, that on the hemisphere is (Vince's Fluxions, p. 277,) .. the resistance on the segment cut off by my is (Vince, p. 277,) and that upon the convex surface of the frustum is .. and this is sufficient to determine the required segment. 738. The sections of a cylinder made by planes parallel to the base are circles. Hence, if R denote the resistance upon the diameter (27) of one of these sections, and a the axis of the cylinder, the resistance upon the semi-surface of the cylinder is (Vince's Flux. p. 277). Also the resistance upon the end of the cylinder when it moves parallel to the axis is 739. Let M be the magnitude of the body, and S and S' the specific gravities of the fluid and the body, respectively; then the descending force is (Vince's Hyd. prop. XVIII) FMXS-MX SM x (S-S') which being constant gives by the question, 740. For the general construction see the note 192, Vol. II, of Newton's Prin. edit. of PP. Le Seur and Jaquier. From this it appears that the resistance (r) upon the curve KA: resist (R) resist(R) upon the base KC:: area of KQHEC: area of the rectangle KI. Now let CI=m, CK = a, CA = ß, KP = x, PH = y', PB=y, KB = s. Then since |