and here s = 400 feet, 0 = 30°, we readily obtain 366. Let / be the common length of the planes, t, t' the times, and v, v' the velocities required; then the velocity acquired Again, V, V' being the velocities acquired and lost by the de scent and ascent in the times t, t', we have V = gt sin. 45° = √ 2gl ( V gt' sin. 30° = √ gl ( √ 2. ‡ 1) 1) whenever the velocities at the end of the motion are respectively v = V" + 2gl. sin. 45° 367. Let the origin of the co-ordinates (x, y, z,) be taken in the intersection of the given line with the plane of the horizon, and the plane of (x, z) be the vertical in which that line is posited; that of (x, y,) being the horizon. Also let (a, b, c,) be the co-ordinates of the given point, and a the inclination of the given line to the horizon. Then drawing any line from the given point to meet the given line in a point whose co-ordinates are (x, y,) it may be considered an inclined plane, whose length and height are easily found to be √√ b2 + (a − x)2 + (e x tan. a) L= 2 = minimum. b2 + ( a − x)2 + c x tan. amin, cx tan. a and putting the differential = 0, we get, after proper deductions, and the solution of a quadratic, x = ccos.ac2-2ac tan. a + (a*+ b2) tan.2 « tan. a which determines the position of the line required. Then A geometrical determination may be made by drawing a line from the given point (P) parallel to the horizon, and meeting the given line in (Q), and taking along the given line (downwards or upwards, according to the position of P) QR = QP, or by drawing a circle touching the lines QP, QR in P, R. This is either obtainable from the construction of the above equation, or directly from the consideration (which may easily be proved) that the times down the chords, originating in the highest or lowest point of a circle any how inclined to the horizon, are equal. 368. The section of the sphere made by a plane passing vertically through the lowest point of the sphere, and another taken any where on its surface, is known to be a circle, whose diameter is the vertical axis of the sphere. Hence, since the times down the chords terminating in the lowest point of a vertical circle are equal to that down its vertical diameter, the time down the line joining the abovementioned points of the sphere is equal to the time down its vertical axis. 369. Let P be the weight of the body, 7 the length of the plane, its inclination to the horizon. Then since the pressure P cos. to the plane P cos. 0, and .. by the question friction = n and the moving force down the plane (not reckoning friction) is P sin. 0, the moving force down it when the retarding influence of friction is considered, will be Hence the force which accelerates the body down the plane is expressed by the positive root of which will give the time required. 370. Let the body be projected in a direction inclined to the horizon at the 0, with the velocity v, and suppose (y, x) the vertical and horizontal co-ordinates, originating in the point of projection, which determine the position of the body at the end of the time t; then since the velocity estimated vertically and horizontally is v sin. 0, v cos. 6, we have the equation to a parabola, whose greatest ordinate, corresponding abscissa, and parameter, are respectively Consequently the equation to the trajectory referred to its vertex, by the horizontal and vertical co-ordinates (w, z) is To resolve the problem, we have by (1) (referring the given coordinates a, b, of the plane, to the point of projection) 4 which gives the direction required; whence also the velocity. 371. Let the equations to the plane (considered a straight line), and parabola described (370), referred to the point of projection, be then, since when the body reaches the plane. y = y', and x = x', we have gx2 2v2 cos.2 0 v cos. § (a-tan. 0) v'cos.2(a-tan 0)2-2bg |