TRACT 38.

PROBLEMS.

325

gives v = 17-134 for the last velocity, when the space s —604, or when the ball arrives at the bottom of the water.

But now to find the time of passing through the water,putting t — any time in motion, and s and v the correspond-ing space and velocity, the general theorem for variable forces

gives t = —. But the above general value of s being x

hyp. log. or 5 x hyp. log. -, therefore its fluxion

s ~ ' v *-a ’ consequently, i or — = fhe correct fluent

which when v = 17-134, ori = 60|, gives 2-6542 seconds,for the time of descent through the water.

A person standing at the distance of 10 feet from the bot-tom of a trail, which is supposed perfectly smooth and hard,desires to know in what direction he must throw an elastic ballagainst it, with a velocity of 80 feet per second, so that, afterreflection from the wall, it may fall at the greatest distance pos-sible from the bottom, on the horizontal plane , which is 2^ feetbelow the hand discharging the ball ?

In the annexed figure let dr r yAbe the ivall against which the

ball is thrown, from the point n.

a, in such a direction, that it ; A \

shall describe the parabolic 1 ® E

curve ae before striking the wall, and afterwards be so re-flected as to describe the curve ep. Now if es be the tan-gent at the point e, to the curve ae described before the re-flection, and ef the tangent at the same point to the curvewhich the ball will describe after reflection, then will the an-gle ref be = ces ; and if the curve pe be produced, so asto have gf for its tangent, it will meet ac produced in e,making bc = ac, and the curve ae will be similar and equal

of which is — x hvn. loo-, i—

- .

x IfLfl) = t the time,

PROBLEM VI.