_Prop._ VIII. The _Lines_ GF, or Times of the Flight of a _Project_ cast with the same Degree of _Velocity_ at different _Elevations_, are as the _Sines_ of the _Elevations_. As _c_ is to _r_; so is _psc/rr_ = GB by the 6 Prop. to _ps/r_ GF; that is, as _Radius_ to _Sine_ of _Elevation_, so the _Parameter_ to the _Line_ GF; so the _Lines_ GF are as the _Sines_ of _Elevation_, and the _Times_ are proportional to those _Lines_; wherefore the _Times_ are as the _Sines_ of _Elevation_: _Ergo constat propositio_. _Prop._ IX. Problem. A _Projection_ being made as you please, having the Distance and Altitude, or Descent, of an Object, through which the Project passes, together with the _Angle_ of _Elevation_ of the _Line_ of _Direction_; to find the _Parameter_ and _Velocity_, that is (in _Fig._ 1.) having the _Angle_ FGB, GM, and MX. _Solution._ As _Radius_ to _Secant_ of FGB, so GM the _Distance_ given to GL; and as _Radius_ to _Tangent_ of FGB, so GM to LM. Then LM - MX in _Heights_, or + MX in _Descents_; or else MX - ML, if the _Direction_ be below the _Horizontal Line_, is the _Fall_ in the _Time_ that the direct _Impulse_ given in G would have carried the Project from G to L = LX = GY; then by Reason of the _Parabola_, as LX or GY, is to GL or YX, so is GL to the _Parameter_ sought. To find the _Velocity_ of the _Impulse_: by Prop. 2, and 4, find the Time in Seconds that a Body would fall the Space LX; and by that dividing the Line GL, the _Quote_ will be the _Velocity_, or Space moved in a Second sought, which is always a mean Proportional between the _Parameter_, and 16 Feet, 1 Inch. _Prop_. X. Problem 2. Having the _Parameter_, Horizontal Distance, and Height or Descent of an _Object_, to find the Elevations of the Line of Direction necessary to hit the given _Object_; that is, having GM, MX, and the greatest _Randon_ equal to half the _Parameter_; to find the _Angles_ FGB. Let the _Tangent_ of the _Angle_ sought be = _t_, the _Horizontal Distance_ GM = _b_, the Altitude of the _Object_ MX = _h_, the _Parameter_ = _p_, and _Radius_ = _r_, and it will be, As _r_ to _t_, so _b_ to _tb/r_ = ML and _tb/r ∓ h_ {in ascents|in descents} = LX, and _ptb/r ∓ ph_ = GL _quad._ = XY _quad. ratione Parabolæ_; but _bb ∓ ttbb/rr_ = GL _quad._ 47. 1. _Euclid_. Wherefore _ptb/r ∓ ph_ = _bb ∓ ttbb/rr_ which Equation transposed, is _ttbb/rr_ = _ptb/r ∓ ph - bb_, divided by _bb_ 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/2h_ ∓ √((_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_)/4_bb_) - 1 or Square of Radius,