412.

Letv be the velocity of projection, v' that after the body has been in motion t", and u the velocity due to its altitude above the horizontal plane after such interval; then, s being the arc described, since

413.

tory, is

By 370, and the question the equation to the trajec

.. putting y = 0, the horizontal range is

v2 √ 3, and consequently the co-ordinates of the point where

2g

the projectile strikes the sonorous body are,

, and its distance from the point of projection is

v2

v2

and

9g

Now, let a denote the velocity of sound as determined by ex

periment; then the time of its moving uniformly through

2v2

9g

Also the time in which the projectile reaches the sonorous body

v2

is that of describing with the uniform velocity v cos. 30°, or 9√3

Hence, m being the whole time given, we have

and by the question, y = 30 feet, x = 120 feet.

v2 = 2g x 80 = 160 g,

.. by substitution and reduction

OSCILLATIONS.

415.

Generally, if t be the time of an oscillation, and I the length of the pendulum, we have (see Translat. of Venturoli, p. 103, or Wood, Prop. 74.)

416.

Let be the length of the cycloidal pendulum; then

the time of an oscillation is

418. Since the moving force here is A B 5 – 3 = 2,

419.

Since the arc of a cycloid 2 x corresponding chord of the generating circle, and the tangent at any point is parallel to that chord, it easily appears that (Fig, in Enunciation)

LV' 30°

.., being the length of the semi-cycloid ABV, or of ABV', we have the time down A'BV' (t) expressed by

ginating in lowest point) is evidently 30°, and the time down it is..

Also the time of oscillation of a pendulum (r) is

421.

.. t: t, :: 2 : π

Let x be the length of the plane, that of the cycloid; then, the elevation of the plane being 30° (see 719),

Also, if I be the length of the common seconds' pendulum, we

have

g

2

the space through which a body would fall in 1" by the force of

gravity.

423.

Let I be the length of the pendulum which loses or gains n' in 24 hours, and .. vibrates 24 × 60 × 60 n times in a day; that of the common pendulum, which vibrates

24 × 60 × 60 times a day. Then, since the number of vibrations for a given time ∞ inversely as the square roots of the lengths, we have 24 × 60 × 60 Fn: 24 × 60 × 60:: √l ≈ ; √ ?

the quantity by which the pendulum ought to be shortened or lengthened, in order to beat true time.