. Differential and integral calculus, an introductory course for colleges and engineering schools. ives /on -d 2 v02 sin $ cos tf> v02 ■ 0 , , „ t 2 v0 sin (3) Range = =-sm2 , when t = Q 9 9 This range will plainly have its greatest value (for a given value of v0) when 4> = - . Hence *V i „_ ._ *„„A * _ V^Vo V2 (3) Maximum range = ■— , when = - and t = To determine the direction in which the particle must be thrown witha given velocity v0 to strike a point A at a distance a from 0, that is, tohave a range a, we have the equations (4) a = — sin 2 , whence 2 = sin-1 —n • 9 v02 §153 APPLICA
. Differential and integral calculus, an introductory course for colleges and engineering schools. ives /on -d 2 v02 sin $ cos tf> v02 ■ 0 , , „ t 2 v0 sin (3) Range = =-sm2 , when t = Q 9 9 This range will plainly have its greatest value (for a given value of v0) when 4> = - . Hence *V i „_ ._ *„„A * _ V^Vo V2 (3) Maximum range = ■— , when = - and t = To determine the direction in which the particle must be thrown witha given velocity v0 to strike a point A at a distance a from 0, that is, tohave a range a, we have the equations (4) a = — sin 2 , whence 2 = sin-1 —n • 9 v02 §153 APPLICATIONS OF INTEGRATION IN KINEMATICS 217 Now when v02 > ag there are two positive angles less than ir, each of which has the sine, —. If we denote one of these angles by 2 h then the other is v — 2 £i. Hence there are two directions in which the particle may bethrown so as to strike A, thesedirections making with the hori- 0i. zontal the angles 0i and Now if we set <fn = 7 —4 + a. That is, the two 2 a, then. = - 0i - -2*4 directions are equally inclined to the direction of maximum range. When Vq2 = ag, sin 2 0 = 1, 0 = j» which gives the maximum range. In that ,case a = 0, which means thatthe two paths coincide at the maximum range. When Vq2 < ag, sin 2 > 1 and 2 0 is imaginary. In that case thepoint A cannot be reached with the initial velocity v0. To find the greatest height to which a particle rises for a given v0 and0, we have only to find the maximum value of y in (2). It is readily found that y is maximum when t = — , and on substituting this Qvalue for t in the equations of (2), we get the coordinates of the highestpoint of the flight. Hence for the point of greatest altitude, 0, we have (5) t = Vq Sin 0 9 v02 sin 0 cos 0 Vo2 . 0 , _- = -sm20, y Q 2<7 snr0. Y Y Directrix D / F 0 X p 0 \ x Since t is half the range, we see thatthe highest point is the mid-point ofthe path. The x and y equation of the pathis obtained by eliminati
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