Corollary (5)

Hence it follows that half the base, or amplitude, of the semi-parabola (which is one-quarter of the entire amplitude) is a mean proportional between its altitude and the sublimity from which a falling body will describe this same parabola.

Given the sublimity and the altitude of a parabola, to find its amplitude.

Let the line ac, in which lie the given altitude cb and sublimity ab, be perpendicular to the horizontal line cd. The problem is to find the amplitude, along the horizontal cd, of the semi-parabola which is described with the sublimity ba and altitude bc. Lay off cd equal to twice the mean proportional between cb and ba. Then cd will be the amplitude sought, as is evident from the preceding proposition.

If projectiles describe semi-parabolas of the same amplitude, the momentum required to describe that one whose amplitude is double its altitude is less than that required for any other.

Let bd be a semi-parabola whose amplitude cd is double its altitude cb; on its axis extended upwards lay off ba equal to its altitude bc. Draw the line ad which will be a tangent to the parabola at d and will cut the horizontal line be at the point e, making be equal to bc and also to ba. It is evident that this parabola will be described by a projectile whose uniform horizontal momentum is that which it would acquire at b in falling from rest at a and whose naturally accelerated vertical momentum is that of the body falling to c, from rest at b. From this it follows that the momentum at the terminal point d, compounded of these two, is represented by the diagonal ae, whose square is equal to the sum of the squares of the two components. Now let gd be any other parabola whatever having the same amplitude cd, but whose altitude cg is either greater or less than the altitude bc. Let hd be the tangent cutting the horizontal through g at k. Select a point l such that hg:gk=gk:gl. Then from a preceding proposition [V], it follows that gl will be the height from which a body must fall in order to describe the parabola gd.

Let gm be a mean proportional between ab and gl; then gm will [Prop. IV] represent the time and momentum acquired at g by a fall from l; for ab has been assumed as a measure of both time and momentum. Again let gn be a mean proportional between bc and cg; it will then represent the time and momentum which the body acquires at c in falling from g. If now we join m and n, this line mn will represent the momentum at d of the projectile traversing the parabola dg; which momentum is, I say, greater than that of the projectile travelling along the parabola bd whose measure was given by ae. For since gn has been taken as a mean proportional between bc and gc; and since bc is equal to be and also to kg (each of them being the half of dc) it follows that cg:gn=gn:gk, and as cg or (hg) is to gk so is to ngmacronsup2 to gkmacronsup2: but by construction hg:gk=gk:gl. Hence ngmacronsup2:gkmacronsup2=gk:gl. But gk:gl=gkmacronsup2:gmmacronsup2, since gm is a mean proportional between kg and gl. Therefore the three squares ng, kg, mg form a continued proportion, gnmacronsup2:gkmacronsup2=gkmacronsup2:gmmacrondotsup2 And the sum of the two extremes which is equal to the square of mn is greater than twice the square of gk; but the square of ae is double the square of gk. Hence the square of mn is greater than the square of ae and the length mn is greater than the length ae.

q. e. d.