TRACT SS.

PROBLEMS.

327

Cor. 1. When b coincides with i, ip will be = bp = be4- ei = 2ei, and the angle ebh will be 45°; as is also ma-nifest front the common modes of investigation.

Cor. 2. When the impetus corresponding to the initialvelocity of the ball is very great compared with ac or bc(fig. 1), then the part ae of the curve will very nearly coin-cide with its tangent, and the direction and velocity at a maybe accounted the same as those at e without any sensibleerror. In this case too the impetus be (fig. 2) will be verygreat compared with bi, and consequently, B and i nearlycoinciding, the angle ebh will differ but little from 45°.

Calcui. From the foregoing construction the calculationwill be very easy. Thus, the first velocity being 80 feetzzw,then (vol. 2 pa. 156 of the Course) if = 80 * f - = 99-48186:= BE the impetus; hence ei = fp = 101‘98186, and bp =be + ei = 201-40372. Now, in the right-angled triangleBiP, the sides Br and bp are known, hence ip = 20T4482,and the angle ibp = 89° 17' 20" : half the suppl. of this angleis 45' 21' 20" = ebh. And, in fig. 1, IP — id = 201-4482— L0 == 191-4482 == dp, the distance the ball falls from thewall after reflection.

problem vii.

From what height above the given point a must an elasticball be suffered, to descend freely by gravity, so that, afterstriking the hard plane at b, it may be reflected back again, tothe point a, in the least time possible from the instant of drop-ping it?

Let c be the point required ; and put ac — .r, andab = a ; then y^/CB = («+.r) is the time in cii,

and ^ fe\ = is the time in ca ; therefore -A

$V(a + .r) —f x is the time down ab, and the timeof rising from b to a again : hence the whole time offalling through cb and returning to a, is iV(a -f- x) b

which must bea min. or 2fl{a-\-x) — fxa minimum,