332

MISCELLANEOUS

TRACT 38,

PROBLEM XI.

If a body begin to move from a, with a given velocity, alongthe quadrant of a circle ab ; it is required to show at whatpoint it will fly off from the curve.

Let d denote the point where thebody quits the circle abd, and then de-scribes the parabola de. Draw the or-dinate df, and iet GAbe the height pro-ducing the velocity at a. Put ga = a,

AC or CD — r, AF = x ; then the velocityin the curve at d will be the same asthat acquired by falling through gf or a -(- x, which is, asbefore, 8\/(a + x) ; but the velocity in the curve is to thehorizontal velocity as nn to mn or as cd to cf by similar tri-angles, that is, as r : r — x : : S\/[x + a) : 8V(x + a) X

r • * , which is to be a constant quantity where the bodyleaves the circle, therefore also (r — x) f(x 4- a) and (?' — xfx (x-f-a) a constant quantity; the fluxion of which made to

... r — 2't

vanish, gives x = —- — af.

Hence, if a = 0, or the body only commence motion at a,then x = -§r, or af Ac when it quits the circle at d. Butif a or ga were = \r or aac, then r — = 0, and the body

would instantly quit the circle at the vertex a, and describea parabola circumscribing it, and having the same vertex A.

PROBLEM XII.

The force of attraction, above the earth, being inversely asthe square of the distance from the centre; it is proposed to de-termine the time, velocity, and other circumstances, attendinga heavy body falling from any given height; the descent at theearth’s surface being 16 T q feet, or 193 inches , in the first se-cond of lime.