. Differential and integral calculus, an introductory course for colleges and engineering schools. it is seen that the position of the directrix is inde-pendent of . Therefore, the parabolic paths of all particles projectedfrom 0 in the same vertical plane and with the same velocity v0 have thesame directrix, whatever be the direction in which they are thrown. Itis to be noted that this common directrix is at a distance above 0 equalto the height to which the body would rise if thrown vertically upwardwith a velocity v0. If the projectile has a high velocity, the resistance of the airso m
. Differential and integral calculus, an introductory course for colleges and engineering schools. it is seen that the position of the directrix is inde-pendent of . Therefore, the parabolic paths of all particles projectedfrom 0 in the same vertical plane and with the same velocity v0 have thesame directrix, whatever be the direction in which they are thrown. Itis to be noted that this common directrix is at a distance above 0 equalto the height to which the body would rise if thrown vertically upwardwith a velocity v0. If the projectile has a high velocity, the resistance of the airso modifies its motion that the foregoing formulae give very inaccurateresults. The law governing the resistance of the air to the motion ofa projectile is not accurately known. Problem 2. To determine the range on an inclined OP be a plane through 0 perpendicular to the plane of the pro-jectiles path and making an angle awith the horizontal plane. We seek todetermine the point P where the pro-jectile pierces this plane. We have already found the equationof the path in the form. tan • x — 1_ 2 v02 cos2 x\ The equation of OP is y = tan a • x. Solving these equations for x and y, we find, after easy reductions, the coordinates of P to be 2 y02 • / x 2v02 cos 0sm (0 — a),y= — — cos sin gcosa gcos a) tan a. §153 APPLICATIONS OF INTEGRATION IN KINEMATICS 219 Substituting these in the equation R = x cos a + y sin a, we have (8) Range on OP = R = 2 y„2 cos sin ( — a). gcosa To determine the value of which makes R a maximum, we setu = cos <£ sin ( — a);D^w = cos cos ( — «) — sin 0 sin (<£ — «) = cos (2 — a). 2 7T ■ Q! 4+ 2 then When D+u = 0, 2 0 - « Then n = cos (| + |) sin(^ - |) = sin2 (| - |) = i[1-C0S(l°£)]=J(1Sina)- Therefore (9) Maximum range on OP = -^ :—— = Itf g COS a g(l + sin a) Setting 0 = - + a in this equation and p for the maximum range, we have (10) _2_ COS0
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Keywords: ., bookcentury1900, bookdecade1910, booksubjectcalculu, bookyear1912
