± √ ¼ pp ± ph - bb / bb , and 'tis this, having made the right Angle GDF, ( Tab. 5. Fig. 3. ) make DF = ½ p , or greatest Range, and GD = b the Horizontal Distance, and DB = h the perpendicular heighth of the Object; to be laid upwards from D, if the Object be above the Horizon, or downwards if below it. Parallel to GD draw FA, and make it equal to GB the Hypothenusal Distance of the Object; and with the Center A and Radius FB = ½ p ± h , sweep an Arch, which shall if the thing be possible, intersect the indeterminate Perpendicular DF in two Points K and L, to which draw the Lines, GL, GK; I say, the Angles DGK, DGL, are the Elevations requisite to strike the Object B.
Demonstration. The Square of FK or FL, is equal to FBq - GBq: or (½p ± h)2 - bb - hh or ¼pp ± ph - bb, and therefore √ ¼pp ± ph - bb is = FK = FL, and by Consequence DK,
DL = ½p ± √ ½pp ± ph - bb . And as DG: DK and DL :: Radius: Tangents sought, which coincides with our Algebraical Expression thereof.
Prop. XI. To determine the Force or Velocity of a Project, in every Point of the Curve it describes.
To do this we need no other Præcognita , but only the third Proposition, viz. That the Velocity of falling Bodies , is double to that which in the same time, would have described the Space fallen by an equable Motion: For the Velocity of a Project, is compounded of the constant equal Velocity of the impressed Motion, and the Velocity of the Fall , under a given Angle , viz. the Complement of the Elevation : For Instance, in Fig. 2. in the time wherein a Project would move from G to L,