TRACT XXXVIII.

MISCELLANEOUS PRACTICAL PROBLEMS, StC, ILLUSTRATINGSOME OF THE FOREGOING PRINCIPLES.

PROBLEM I.

It is required to find the Diameter of a Circular Parachute ,by means of which a man of \50lb weight may descend to theearth, from a Balloon at a height in the air, with the Velocityof only 10 feet in a second of time, being the Velocity acquiredby a body freely descending through a space of only 1 foot 6|-inches, or of a man jumping down from a height of 18|- inches:the Parachute being made of such materials and thickness, thata circle of it of SO feet diameter, weighs only 150 lb, and so inproportion more or less according to the area of the circle.

If a falling body descend with a uniform velocity, it mustnecessarily meet with a resistance, from the medium it de-scends in, equal to the whole weight that descends. Let xdenote the diameter of the parachute, and a — '7854; thenax z will be its area, and as 50 z : x z : : 150 : f 6 x z the weightof the same, to which adding 150lb, the man’s weight, thesum 150 will be the whole descending weight. Again,

in the table of resistances, at pa. 189, Tract 36, art. 42, wefind that a circle of of a square foot area, moving with 10feet velocitj’, meets with a resistance of *57 ounces= - 0475lb;and the resistances, with the same velocity, being as the sur-faces, therefore as : '0475 : : ax z : • 2VA75ax z zz •16788x*the resistance of the air to the parachute, to which the de-scending weight must be equal; that is, T6788.F 2 = jfx z +150 ; hence T0788x 2 = 150, or x z = ISSIO’S, and hence x =37 feet, the diameter of the parachute required.