CASE 2. If the velocity in the ascent of the body be expounded by the length AP as before, and the resistance be made as AP2 + 2BAP, and if the force of gravity be less than can be expressed by DA2; take BD of such a length, that AB2 - BD2 may be proportional to the gravity, and let DF be perpendicular and equal[Pg 282] to DB, and through the vertex F describe the hyperbola FTVE, whose conjugate semi-diameters are DB and DF, and which cuts DA in E, and DP, DQ in T and V; and the time of the whole ascent will be as the hyperbolic sector TDE. For the decrement PQ. of the velocity, produced in a given particle of time, is as the sum of the resistance AP2 + 2BAP and of the gravity AB2 - BD2, that is, as BP2 - BD2. But the area DTV is to the area DPQ as DT2 to DP2; and, therefore, if GT be drawn perpendicular to DF, as GT2 or GD2 - DF2 to BD2, and as GD2 to BP2, and, by division, as DF2 to BP2 - BD2. Therefore since the area DPQ is as PQ, that is, as BP2 - BD2, the area DTV will be as the given quantity DF2. Therefore the area EDT decreases uniformly in each of the equal particles of time, by the subduction of so many given particles DTV, and therefore is proportional to the time. Q.E.D. CASE 3. Let AP be the velocity in the descent of the body, and AP2 + 2BAP the force of resistance, and BD2 - AB2 the force of gravity, the angle DBA being a right one. And if with the centre D, and the principal vertex B, there be described a rectangular hyperbola BETV cutting DA, DP, and DQ produced in E, T, and V; the sector DET of this hyperbola will be as the whole time of descent. For the increment PQ of the velocity, and the area DPQ proportional to it, is as the excess of the gravity above the resistance, that is, as BD2 - AB2 - 2BAP - AP2 or BD2 - BP2. And the area DTV is to the area DPQ as DT2 to DP2; and therefore as GT2 or GD2 - BD2 to BP2, and as GD2 to BD2, and, by division, as BD2 to BD2 - BP2. Therefore since the area DPQ is as BD2 - BP2, the area DTV will be as the given quantity BD2. Therefore the area EDT increases uniformly in the several equal particles of time by the addition of as many given particles DTV, and therefore is proportional to the time of the descent. Q.E.D. [Pg 283] COR. If with the centre D and the semi-diameter DA there be drawn through the vertex A an arc At similar to the arc ET, and similarly subtending the angle ADT, the velocity AP will be to the velocity which the body in the time EDT, in a non-resisting space, can lose in its ascent, or acquire in its descent, as the area of the triangle DAP to the area of the sector DAt; and therefore is given from the time given. For the velocity in a non-resisting medium is proportional to the time, and therefore to this sector; in a resisting medium, it is as the triangle; and in both mediums, where it is least, it approaches to the ratio of equality, as the sector and triangle do. SCHOLIUM.