382.

Let the inclination of the plane, passing through the centre of the circle, and tangent at its extremity, be ; then s being the distance of the point of projection from the tangent, v the velocity of projection, v' that with which the body is reflected by the tangent plane, t the time from the hand to the tangent, and t' that from the tangent to the hand, we easily obtain

v'

sin. 0

But (velocity with which the body strikes the tangent

and making r — s the radius vector (since it is the man's distance from the centre of the circle), the polar equation to his locus, is

If it were required to determine the curve by rectangular coordinatės, we have

PROJECTILES IN A VACUUM.

If it were required to determine the curve by rectangular co

ordinates, we have s = r + √x2 + y2, sin. 0 =

y

which

√x2 + y2

being substituted would give the equation..

In both cases the equation will be very complex. It would be rather laborious than difficult to rationalize these results.

383.

Let a be the altitude through which the body falls to acquire the velocity of projection v, and suppose that after describing with this velocity the arc s of the trajectory, it is reflected by a hard normal plane, at the point whose vertical and horizontal co-ordinates, originating in the point of projection, are (y, x); then it being again projected by the intervention of this plane with a certain velocity v', that plane (by the question) must be so situated as to reflect the body precisely to a point whose co-ordinates (-y', x) measured from the point of reflection are a — y, x. Now, by (370), we have

the equation to the first parabola.

Again, since the horizontal velocity of a projectile is uniform, we have (t being the time of describing a with the velocity v).

Hence, and by equation (1), since the body is reflected with half the velocity of compact,

Also, since the body, after reflection, is projected in the direc

tion of the tangent at the point (x, y), we have the inclination (0) of that direction to the horizon expressed by

dy x

tan. 0 =

Hence, since is also the inclination of the normal plane to the vertical, we readily obtain its equation, referred to the point of projection by y', x", viz.,

determining the position required.

Again, t, t,, t, being the times of describing a from rest, and x with the uniform horizontal velocities v, v', we have

384. By (374) the equation to the locus of the centre of

2 x 49 cos.' 45° × 16

PROJECTILES IN A VACUUM.

the equation to a parabola whose parameter and greatest ordinate

And that of the plane referred also to the point of projection, is

y' = x'√3a.

where a is the distance of the point of projection from the plane's intersection with the vertical.

Now, when the body strikes the plane, we have, x = x', y = y', which give

by the resolution of which the time required may be found.

386.

Let be the required, and a the given velocity; then (370) the greatest ordinate and corresponding abscissa being

387.

Let x be the distance of the point of concourse from

the bottom of the vertical line; then

But here y is negative, 0 = 0, and v2 = 2ga (a being the altitude of the tower).

389.

Generally, if A, B; a,b,; 0, o denote the weights of the bodies, the velocities of projection, and directions respectively, then we have for the time (t), (x, y ; x', y'; x′′, y") denoting the horizontal and vertical co-ordinates of the trajectories described by the projectiles and their centre of gravity,

x= ta x cos. 0, x = tb x cos.

y=ta × sin.0 - 2. t2, y′ = tb × sin. o̟

2