is tt ⁄ rr = pt ⁄ br ∓ ph ⁄ bb - 1.
this Equation shews the Question to have 2 Answers, and the Roots thereof are t⁄r = p⁄2b ∓ √
pp ∓ 4ph / 4bb - 1; from which I derive the following Rule.
Divide half the Parameter by the Horizontal distance, and keep the Quote; viz. p⁄2b then say, as square of the distance given to the half Parameter, so half Parameter ∓ double
height | descent to the square of a Secant =
pp ∓ 4ph / 4bb. The Tangent answering to that Secant, will be √
pp ∓ 4ph / 4bb - 1 or Square of Radius, so then the sum and difference of the afore-found Quote, and this Tangent will be the Roots of the Equation, and the Tangents of the Elevations sought.
Note here, that in Descents, if the Tangent exceed the Quote, as it does when ph is more than bb, the direction of the lower Elevation will be below the Horizon, and if ph = bb, it must be directed Horizontal, and the Tangent of the upper Elevation will be pr⁄b: Note likewise, that if 4bb + 4ph in Ascents, or 4bb - 4ph in Descents, be equal to pp, there is but one Elevation that can hit the Object, and its Tangent is pr⁄2b. And if 4bb + 4ph in Ascents, or 4bb - 4ph in descents, do exceed pp, the Object is without the reach of a Project cast with that Velocity, and so the thing impossible.
From this Equation 4 bb ∓ 4 ph = pp are determined the utmost limits of the reach of any Project , and the Figure assigned, wherein are all the heights upon each Horizontal