PROBLEM. Fig. 1. To describe the path of a projectile. Draw AL the horizontal plane, and, from a scale of equal párts, lay off the amplitude AX, and through the point A erect a perpendicular AB cqual to the impetus taken from the same {cale ; through B draw the directrix parallel to AK; then bisect AK in Q, and draw QN at right angles to AK; upon A, as centre with the distance AB, describe the semicircle BFFR, and the point F is the focus. Or, If the direction AD is given, upon AB, as diameter, describe a semicircle BDA; and through the point of intersection D draw BD, and produce it to F; fo fhall BD and DF be equal, and F will be the focus. Or, Through the point D draw PD parallel to the horizon ; then Thall PD=DH, and NH=HF, and H will be the vertex. Cor. 4. times PD is equal to the amplitude. PROBLEM I. The impetus of a piece and the angle of elevation being given, to find the amplitude. EXAMPLE I. How far will a cannon, whose impetus is 1200 feet, carry, at an elevation of 30°? Geometrically. Let AB represent the impetus of the piece, or the velocity a heavy body would acquire in falling from B to A. Through the point A draw the horizontal line AL, and make the angle O. LAM LAM equal to the angle of elevation. From the centre A, with the radius AB, describe the semicircle BFOER; its circumference shall be the locus of the foci of all the parabolas that can be described by a projectile thrown from A, with the velocity it could acquire in falling from B to A ; for, by a known property of the parabola, the distance of the focus from A is always equal to ône-fourth of the parameter of the diameter that passes through A, that is, to AB; all the foci must, therefore, be found in the femicircle BFCFR. It will therefore be casy to determine the parabolas, when the direction of the projectile is given ; for if, upon the impetus AB, you describe a semicircle BDIA, you need only join BD, and lay off BD equal to DF, and F will be the focus; and if through F you draw the line QF perpendicular to the horizontal line AL, it shall be the axis ; and H, the middle point between F and N, shall be the vertex of the parabola. 4xFH is the length of the parameter of the axis. If a line HP be drawn through the point H perpendicular to AB, the straight line BF and PH will bisect each other; also AM, the line of direction, will pass through the point of intersection in D, and bifect the line BF at right angles; and therefore the semicircle BDdA will pass through the same point D. The amplitude of any parabola is equal to four times the line of twice the complement of the angle of elevation : PD is the line of the angle PCD, and the angle PCD is twice the angle PAD, because the one is at the centre and the other at the circumserence; but the angle PAD is the complement of the angle of elevation DAK; therefore PD is the fine of twice the complement of the angle of elevation; and 2PD is equal to PH; but 2PH is equal to AK; thereforé AK is equal to 4PD. Hence it will follow, that when the angle of elevation becomes 45°, the points F and shall ful in the point 0, and AK becomes twice the impetus. The fine PD is the co-fine of double 1 double 45', which is the fine of 90°, or the radius; and, as the fine of 90° is the greatest, we may infer, that if a body is projected with an elevation of 45°, it will be carried farther on the horizontal plain, than, if projected with the fame velocity, in any other direction. Also, If of two directions the elevation of the one exceeds 45° as much as the elevation of the other wants of 45°, their amplitudes will be equal, for the angles are complements of each other, and the fines of double of thefe angles must be equal, because they are supplements to two right angles to one another, but the amplitudes of the parabola is always quadruple of these fines, and therefore they must also be equal, Let the impetus be 3600, and the angle of elevation 75°. rethe amplitude. As radius 90 10.00000 3.85733 9.69897 3.55639 From the preceding example, it is evident, that the impetus of a piece is equal to the amplitude, when fired off at the angles of 15° or 75° By Scale and Compasses. In Ex. I. Extend the compaffes from the radius to the fine of 60°, the same extent will reach from 2400 on the line of numbers, to 2078, the amplitude required. PROBLEM II. The amplitude and impetus being given, to find the elevation. EXAMPLE I. At what elevation will a mark be hit, distant 5100 yards, the impetus being 3000? Ex. 2. At what elevation will a mark be hit, distant 180. yards, the impetus being 900 yards ? Anf. 45° Ex. 3. At what angle will an object be hit, distant 420 yards, the impetus being being 4000? $ 15° 50' lowest. 274° 10' highesta By Scale and Compaffes. The extent from twice the impetus on the line of numbers, to the amplitude, will reach from the radius on the lineof fines, to the line of double the elevation. PROBLEM PROBLEM III. Given the amplitude and the angle of elevation, to find the i npetusa EXAMPLE I. What impetus will carry a ball 3520 yards, at an elevation of 30° or of 60° ? The extent from twice the angle of elevation on the line of fines, will reach from į amplitude; on the line of numbers, to the impetus. Ex. 2. The amplitude is 3000, and the direction 45°, required the impetus. Anf. 1500. Ex. 3. The amplitude is 5200, and elevation 75, required Anf. 5200. the impetus. PROBLEM IV. I he amplitude and direclign being given, to find the height of the projection. EXAMPLE. I. The amplitude being 1200 yards, and elevation 30°, required the height of the projection. As |