EXAMPLE.

A cannon being fired at an angle of 24° 5', the greatest altitude of the ball 180 yards; another was fired off, and the greatest altitude of the ball was 400 yards; at what angle of elevation was the cannon fired off the second time?

As the first altitude 180 =

2.25527

Is to the verfed fine of twice the first elev. 48° 20'

4.52249

The extent from 180, on the line of numbers, to 400, will reach from 48° 10' to 74° 56' on the line of verfed fines

Ex. 2. A ball was projected at an angle of 40° 30', its greateft altitude being 500 yards; afterwards another was projected, whose altitude was 400, required the elevation of the piece. Anf. 35° 31'. Ex. 3. The greatest elevation of a ball being 450 yards, the elevation 36° required the elevation of another projection, the greatest altitude being 240 yards. Anf. 25° 25′

PROBLEM VIII.

The elevation and amplitude being given, to find the time of the

flight.

EXAMPLE.

How long will a ball, fired off at an angle of 58°, remain in

the air, the amplitude being 528c feet?

As

This Problem is neceffary in adjusting the fufce of bombs, which are generally fired off at an angle of 45°.

It is common among gunners to find the angle between the object and the zenith, and take the complement of half that angle for their elevation. And because a lefs charge of powder will ferve with this elevation than with any other, they find, by trial, what charge will reach the object,

PROBLEM IX.

The amplitude of the projectile, with a given charge of powder being given, to find what charge of powder will be neceffary to hit an object at any other diftance, (not greater than the outmoft range) the elevation being the fame.

EXAMPLE.

If 16 lb. of powder will foot a cannon ball to the distance of 6oco yards, required the neceflary charge to fhoot the fame ball 5000, with the fame elevation.

Or fay, numerically,

As 6000: 16: 5000: 13.7 lb.

PROBLEMS on Afcents and Defcents.

A projectile thrown on an afcent, with the velocity it would acquire in falling from B to A, in the direction AE, will strike the line AN in K, fo that AK will be equal to 4CD. Suppofing the angle KAG a right angle, the angles GAB=GBA, and that a femicircle on G, as centre with the radius GB, cuts the line of direction in D, and that DC is parallel to AN, meeting AB in the point C,

Because the angles KAD, ADC, are equal, being the alternate angles, and AK touches the circle, and AD cuts it, the angles KAD, DBA, are equal; therefore the angle DBA= CDA, confequently the triangles ACD, ADB, are fimilar, having the angle at A common; therefore AC: AD: : AD: AB. Again: Because the triangles ACD, PAK, are fimilar, AP : PK :: PK: 4AB; therefore AD=PK, consequently CD=AK,

Cor 1. Through D draw a line parallel to AB, cutting the circle in Dd, and join AD, then will the projectile, thrown in the direction Ad, ftrike the line NA in the point k, for CD= cd=AB=AK.

Cor. 2. Parallel to AB draw HL, a tangent to the circle in H, join AH, then fhall AH be the direction which shall carry the projectile fartheft on the line AN; because, when D coin

cides with H, CD is the greateft poffible, and confequently AK (4CD) is the greatest distance the projectile can be carried to, with the velocity acquired in falling from B to A.

Cor. 3. It is plain that the angle HAN=HBA=HAB; therefore the direction AH bifects the angle BAN.

Cor. 4. The lines AD, Ad, make equal angles with AR, confequently the angles DAN, dAN, are equal, and the diftance AK is invariably the fame.

Cor. 5. When AK is given, and the direction required, take AR=AK, and through RD parallel to AB, meeting the cir

cle in D, d, draw AD, Ad, and thefe will be the directions.

PROBLEM I.

The horizontal diftance, and the perpendicular height of the object above the level of projection, alfo the impetus being given, to find

the elevations.

EXAMPLES.

Let the horizontal diftance be 7000, impetus 4200, and the horizontal height 744, required the directions.

As the horizontal distance 700c =

Is to radius 90

So is the height of the object 744

To tangent angle of obliquity 6° 4′ =

Half of which, added to 45°, makes 48° 2′.

Ax: AK: AC AG, that is,

7000: 7040 :: 2100 :: 2112.

3.84510

10.00000

2.87157

9.02047

Half of which added to or subtracted S 58° 26' higher.

from 48° 2', gives

37° 38 lower.

PROBLEM II.

Given the angles of direction, obliquity of the plane, and amplitude, to find the impetus.

As fine ang. dAzx 231° 34′ = 9.71891 into fine ang. BAD, S 52° 22' = 9.89869

Is to the fquare of

19.61760

the fine of BAZ. J 83° 56′ = 9.99756 19.99512

The angles of direction, obliquity of the plane, and impetus being given, to find the random.

EXAMPLE