{VERSION 4 0 "IBM INTEL NT" "4.0" } {USTYLETAB {CSTYLE "Maple Input" -1 0 "Courier" 0 1 255 0 0 1 0 1 0 0 1 0 0 0 0 1 }{CSTYLE "2D Math" -1 2 "Times" 0 1 0 0 0 0 0 0 2 0 0 0 0 0 0 1 }{CSTYLE "2D Input" 2 19 "" 0 1 255 0 0 1 0 0 0 0 0 0 0 0 0 1 } {CSTYLE "2D Output" 2 20 "" 0 1 0 0 255 1 0 0 0 0 0 0 0 0 0 1 } {CSTYLE "" -1 256 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 257 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 258 "" 1 16 0 0 0 0 0 1 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 259 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 260 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 1 } {CSTYLE "" -1 261 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 262 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 263 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 264 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 265 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 1 } {CSTYLE "" -1 266 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 267 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 268 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 269 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 270 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 1 } {CSTYLE "" -1 271 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 1 }{PSTYLE "Normal " -1 0 1 {CSTYLE "" -1 -1 "Times" 1 12 0 0 0 1 2 2 2 2 2 2 1 1 1 1 }1 1 0 0 0 0 1 0 1 0 2 2 0 1 }{PSTYLE "Maple Output" -1 11 1 {CSTYLE "" -1 -1 "Times" 1 12 0 0 0 1 2 2 2 2 2 2 1 1 1 1 }3 3 0 0 0 0 1 0 1 0 2 2 0 1 }{PSTYLE "Maple Output" -1 12 1 {CSTYLE "" -1 -1 "Times" 1 12 0 0 0 1 2 2 2 2 2 2 1 1 1 1 }1 3 0 0 0 0 1 0 1 0 2 2 0 1 }{PSTYLE "Maple Plot" -1 13 1 {CSTYLE "" -1 -1 "Times" 1 12 0 0 0 1 2 2 2 2 2 2 1 1 1 1 }3 1 0 0 0 0 1 0 1 0 2 2 0 1 }} {SECT 0 {EXCHG {PARA 0 "" 0 "" {TEXT 258 11 "Cannon ball" }}{PARA 0 " " 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 173 "We consider the p roblem of firing cannon balls: this involves two-dimensional motion in a gravitational field including air resistance. We set up Newton's eq uations for the " }{TEXT 256 1 "x" }{TEXT -1 5 " and " }{TEXT 257 1 "y " }{TEXT -1 26 " components of the motion." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 259 1 "g" }{TEXT -1 40 " is the gravit ational acceleration, and " }{TEXT 260 1 "b" }{TEXT -1 27 " the air dr ag coefficient. " }{TEXT 263 1 "m" }{TEXT -1 80 " is the mass of the b all. We assume the drag to be proportional to the velocity." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 145 "We set up Newt on's equations as first-order differential equations for the momentum \+ and position vector components (Hamilton's equation). In the " }{TEXT 262 1 "x" }{TEXT -1 56 " direction we have drag as the only force, whi le in the " }{TEXT 261 1 "y" }{TEXT -1 32 " direction gravity acts as \+ well." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart;" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "Momx:=m*diff(vx(t),t)=-b*vx( t);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%MomxG/*&%\"mG\"\"\"-%%diffG6 $-%#vxG6#%\"tGF/F(,$*&%\"bGF(F,F(!\"\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 35 "Momy:=m*diff(vy(t),t)=-m*g-b*vy(t);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%MomyG/*&%\"mG\"\"\"-%%diffG6$-%#vyG6#%\"tGF/F(,&* &F'F(%\"gGF(!\"\"*&%\"bGF(F,F(F3" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 60 "The initial conditions are specified through the parameters:" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 44 "IC:=vx(0)=v0*cos(theta),vy(0 )=v0*sin(theta);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#ICG6$/-%#vxG6# \"\"!*&%#v0G\"\"\"-%$cosG6#%&thetaGF-/-%#vyGF)*&F,F--%$sinGF0F-" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 42 "sol:=dsolve(\{Momx,Momy,IC\} ,\{vx(t),vy(t)\});" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$solG<$/-%#vxG 6#%\"tG*(%#v0G\"\"\"-%$cosG6#%&thetaGF--%$expG6#,$*&*&%\"bGF-F*F-F-%\" mG!\"\"F:F-/-%#vyGF),$*&,&*&F9F-%\"gGF-F-*&F2F-,&FAF-*(F,F--%$sinGF0F- F8F-F-F-F:F-F8F:F:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "assig n(sol);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 81 "We can verify that the found solutions indeed satisfy the differential equations:" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "simplify(Momx);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/,$**%#v0G\"\"\"-%$cosG6#%&thetaGF'%\"bGF'-%$ex pG6#,$*&*&F,F'%\"tGF'F'%\"mG!\"\"F5F'F5F$" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 9 "evalb(%);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#%%trueG " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "simplify(Momy);" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#/,$*&-%$expG6#,$*&*&%\"bG\"\"\"%\"tGF- F-%\"mG!\"\"F0F-,&*&F/F-%\"gGF-F-*(%#v0GF--%$sinG6#%&thetaGF-F,F-F-F-F 0F$" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "evalb(%);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#%%trueG" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 56 "It is of interest to Taylor expand the solutions around " }{TEXT 264 1 "t" }{TEXT -1 76 "=0. This shows how the velocities begin to dev iate from their inital values:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "taylor(vx(t),t=0,3);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#++%\"t G*&%#v0G\"\"\"-%$cosG6#%&thetaGF'\"\"!,$*&*(F&F'F(F'%\"bGF'F'%\"mG!\" \"F2F',$*&*(F&F'F(F')F0\"\"#F'F'*$)F1F7F'F2#F'F7F7-%\"OG6#F'\"\"$" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "taylor(vy(t),t=0,3);" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#++%\"tG*&%#v0G\"\"\"-%$sinG6#%&thetaGF '\"\"!,$*&,&*&%\"mGF'%\"gGF'F'*(F&F'F(F'%\"bGF'F'F'F1!\"\"F5F',$*&*&F4 F'F/F'F'*$)F1\"\"#F'F5#F'F;F;-%\"OG6#F'\"\"$" }}}{EXCHG {PARA 0 "" 0 " " {TEXT -1 215 "The change is linear in time, and depends in the drag \+ part on the initial velocity components, while the contributions from \+ gravity are obvious in the linear term. Note that the quadratic terms \+ are more complicated." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 192 "It is of interest to investigate various aspects of t he solution. The most common problem concerns the dependence of the tr ajectory on the inclination of the cannon for fixed drag coefficient \+ " }{TEXT 265 1 "b" }{TEXT -1 99 ". Note that in real life one will inc lude additional factors, such as the wind speed and direction." }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 88 "In order \+ to plot solutions we need to specify the parameters. In SI (MKSA) unit s we have" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "g:=9.81;" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"gG$\"$\")*!\"#" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 23 "we pick a mass of 1 kg:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 5 "m:=1;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"mG\" \"\"" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 29 "a drag coefficient (in kg /s):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 7 "b:=0.5;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"bG$\"\"&!\"\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 6 "vx(t);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#*(%#v0G\"\" \"-%$cosG6#%&thetaGF%-%$expG6#,$*&*&%\"bGF%%\"tGF%F%%\"mG!\"\"F3F%" }} }{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "simplify(%);" }}{PARA 11 " " 1 "" {XPPMATH 20 "6#*(%#v0G\"\"\"-%$cosG6#%&thetaGF%-%$expG6#,$%\"tG $!+++++]!#5F%" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "simplify(v y(t));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,($!++++i>!\")\"\"\"*&$\"+++ +i>F&F'-%$expG6#,$%\"tG$!+++++]!#5F'F'*(%#v0GF'-%$sinG6#%&thetaGF'F+F' F'" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 170 "Suppose we fix now the ini tial speed (by fixing the amount of powder for the firing of the canno n). We can still vary the inclination angle. How can we achieve this g oal?" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 27 "A gain in SI units (in m/s):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 7 "v0:=50;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#v0G\"#]" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 52 "We can choose a parameter for the inclina tion angle:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "simplify(vy( t));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,($!++++i>!\")\"\"\"*&$\"++++i >F&F'-%$expG6#,$%\"tG$!+++++]!#5F'F'*(%#v0GF'-%$sinG6#%&thetaGF'F+F'F' " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 74 "To generate functions rather \+ than expressions we need the unapply command:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 37 "Vx:=unapply(simplify(vx(t)),theta,t);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#VxGR6$%&thetaG%\"tG6\"6$%)operatorG%&arro wGF),$*&-%$cosG6#9$\"\"\"-%$expG6#,$9%$!+++++]!#5F3$\"#]\"\"!F)F)F)" } }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 37 "Vy:=unapply(simplify(vy(t) ),theta,t);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#VyGR6$%&thetaG%\"tG6 \"6$%)operatorG%&arrowGF),($!++++i>!\")\"\"\"*&$\"++++i>F0F1-%$expG6#, $9%$!+++++]!#5F1F1*(%#v0GF1-%$sinG6#9$F1F5F1F1F)F)F)" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 134 "We are ready to integrate the equations for th e position vector components. A dummy variable has to be used for the \+ time integration (" }{TEXT 267 1 "s" }{TEXT -1 11 " replacing " } {TEXT 266 1 "t" }{TEXT -1 60 "). Again we use the unapply procedure to generate a mapping." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 44 "X:=unapply(int(Vx(theta,s),s=0..t),theta, t);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"XGR6$%&thetaG%\"tG6\"6$%)op eratorG%&arrowGF),&*&-%$cosG6#9$\"\"\"-%$expG6#,$9%$!+++++]!#5F3$!$+\" \"\"!*&$\"$+\"F>F3F/F3F3F)F)F)" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "X(0.5,1);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+<$=IX$!\")" }} }{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 44 "Y:=unapply(int(Vy(theta,s), s=0..t),theta,t);" }}{PARA 12 "" 1 "" {XPPMATH 20 "6#>%\"YGR6$%&thetaG %\"tG6\"6$%)operatorG%&arrowGF),,9%$!++++i>!\")*&$\"++++CRF1\"\"\"-%$e xpG6#,$F.$!+++++]!#5F5!\"\"*($\"$+\"\"\"!F5-%$sinG6#9$F5F6F5F=$\"++++C RF1F5*&$F@FAF5FBF5F5F)F)F)" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "Y(0.5,1);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+&>m$o9!\")" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 108 "We have functions that generate t he position vector for given inclination angle theta (in radians) and \+ time " }{TEXT 268 1 "t" }{TEXT -1 78 " (in seconds). Thus, we can plot the trajectory parametrically (the help page " }{TEXT 19 5 "?plot" } {TEXT -1 41 " includes an example for this plot mode):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 57 "plot([X(0.5,t),Y(0.5,t),t=0..10],x= 0..X(0.5,10),y=0..20);" }}{PARA 13 "" 1 "" {GLPLOT2D 642 230 230 {PLOTDATA 2 "6%-%'CURVESG6$7U7$$\"\"!F)F(7$$\"35'p\"**>vBaY!#<$\"35q([ [Y/a[#F-7$$\"3q!e/&)pQ;1*F-$\"3Oonfd=fDZF-7$$\"35Y!**Q4)eq7!#;$\"3s$)4 V;MS&['F-7$$\"3?h^b\\n8=;F8$\"3GKkLWjcx!)F-7$$\"3cwrf+!o@M#F8$\"3?Rkv# Qn&36F87$$\"37H5iXo/(*HF8$\"3%=VW_Z0zL\"F87$$\"3)e,PI*oh#e$F8$\"3u.8vW :R+:F87$$\"3/M2/[IPsSF8$\"3eApt6LC)f\"F87$$\"38E26b2(3`%F8$\"3k,rE`cE^ ;F87$$\"35QakD$*3e\\F8$\"3')Q%4a3\\%f;F87$$\"3VbDRvP9T`F8$\"32uxb!)H4D ;F87$$\"3GY=k$)=5&p&F8$\"3?z&zv#e&*\\:F87$$\"3%eA9z(p_wfF8$\"39U2hCEj` 9F87$$\"31+Op2RqiiF8$\"3KDcT\\av98F87$$\"33g)GUNB1_'F8$\"3;f>Fe&og9\"F 87$$\"3:nfRM*)4WnF8$\"3Ki))H_R%fe*F-7$$\"3!pA?`j!zFpF8$\"3'RpP871Bp(F- 7$$\"33(om;#ysCrF8$\"3aDYn&)*)fVHo(F8$!3%z1*R=rl=aF-7$$\"3]xA#z!))f*y(F8$!3OA&pAqY&)Q)F-7$$ \"3_D3*p[-)*)yF8$!3KI&3*\\'Q(f6F87$$\"3ee(4([x#G(zF8$!3'4xsLJWLY\"F87$ $\"3+f?ip>o`!)F8$!3]`))=1nY*z\"F87$$\"3d>uQ%3e!H\")F8$!3A)f!4Ic:d@F87$ $\"3q,9P=\"Q#)=)F8$!3)[g80x<\\Z#F87$$\"3u8-zJp3Y#)F8$!3P_MVIq8CGF87$$ \"3)e:VY!f()*H)F8$!3AEYi:+&3>$F87$$\"3+))>VC(3sM)F8$!3Q5['oih[b$F87$$ \"35kH'p8?&)Q)F8$!3G1K$on<:\"RF87$$\"3;A()*QTT(H%)F8$!3UDR2C497VF87$$ \"3gWqrXX-j%)F8$!3=ZOA4,&en%F87$$\"3G*Rj?$f.&\\)F8$!3Gd;?I'*on]F87$$\" 3V$zheF87$$\"3i#4 a9BN*o&)F8$!3wIn&zrIF>'F87$$\"3SH#pv!Hf*e)F8$!3oK*p@M#)\\e'F87$$\"3_NS &*RRz2')F8$!3_&RRB`i/(pF87$$\"3q\\Nyid%\\i)F8$!3=.oia]\"fP(F87$$\"3e\\ =8mx!)R')F8$!3St9p\"Rg!oxF87$$\"3KHS'RU$\\`')F8$!3QmO7+')eq\")F87$$\"3 QOqW@aql')F8$!3#zX=S3N6d)F87$$\"3kp.;9#\\en)F8$!3f0r,)Rm-%*)F87$$\"3G@ zbP7L'o)F8$!3K:Q%R%*oWO*F87$$\"3![%>h!*=x%p)F8$!3,kXe?#4[u*F87$$\"3Y*y f\"pj*Gq)F8$!3Zz%yh0?^,\"!#:7$$\"3CY\\ho'3*4()F8$!3-%=%=yF4a5Fgz7$$\"3 8hf5rXp;()F8$!3AO:nv([g4\"Fgz-%'COLOURG6&%$RGBG$\"#5!\"\"F(F(-%+AXESLA BELSG6$Q\"x6\"Q\"yF]\\l-%%VIEWG6$;F($\"+rXp;()!\");F($\"#?F)" 1 2 0 1 10 0 2 9 1 4 2 1.000000 45.000000 45.000000 0 0 "Curve 1" }}}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 158 "The limits on the axes were adjusted by \+ hand. If we wish to automate the process we can do this by finding the times at which the ball intercepts the surface." }}{PARA 0 "" 0 "" {TEXT -1 19 "This is done below." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 114 "As an exercise the reader may want to pl ay around with the amount of friction. For a very small friction const ant " }{TEXT 270 1 "b" }{TEXT -1 71 " the parabolic shape known from t extbook solutions should be recovered." }}{PARA 0 "" 0 "" {TEXT -1 0 " " }}{PARA 0 "" 0 "" {TEXT -1 47 "The intercept with the surface is obt ained from" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "t0:=solve(Y(0 .5,t)=0,t);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#t0G6$$\"+(GJ$oP!\"* \"\"!" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 19 "Note that we cross " } {TEXT 271 1 "y" }{TEXT -1 285 "=0 at the beginning of the motion as we ll. The order in which the two solutions are found may vary. It does n ot matter for the parametric plot which way the trajectory is traced o ut. We do not need to specify the range on the axes, as they are autom atically adjusted with the solution." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 41 "plot([X(0.5,t),Y(0.5,t),t=t0[1]..t0[2]]);" }}{PARA 13 "" 1 "" {GLPLOT2D 647 259 259 {PLOTDATA 2 "6%-%'CURVESG6$7W7$\"\"!F (7$$\"1$*)3$))[r$y\"!#:$\"1(>03j4Bm*!#;7$$\"1`6n&>v6`$F,$\"1j.ky^W'*=F ,7$$\"1i+%4,4F-&F,$\"1iKfqs3xEF,7$$\"1aBHD+$y['F,$\"17B1K)*[JMF,7$$\"1 -=Uy7FL$\"1(4BWw[A_'F,7$$\"1DE+`)zUd \"FL$\"106FLQd\")yF,7$$\"13H$G1Q\"Q=FL$\"1a[[)3G\"Q!*F,7$$\"1IK\"z%[;, @FL$\"1:K:(*4X85FL7$$\"1]ww7EqiBFL$\"1f**zPGU;6FL7$$\"1Us'\\U-Kh#FL$\" 1J$3k:K*37FL7$$\"1&*4setmgGFL$\"1Ra=$R^RH\"FL7$$\"1)z1\\L!RqIFL$\"1BgC %>Y1O\"FL7$$\"1At9vZg(H$FL$\"1W%>]1.pU\"FL7$$\"1]AwP#[m^$FL$\"14#HOx(Q %[\"FL7$$\"1yAZKhW>PFL$\"1bC$HE[:`\"FL7$$\"1*e:^$H#o*QFL$\"1#p'**))ejn :FL7$$\"17:B<1m*4%FL$\"1G7;$3?Dg\"FL7$$\"1i)zP,z[E%FL$\"1v0Y=K]D;FL7$$ \"1oAd&)ot\\WFL$\"1c[2cd&\\k\"FL7$$\"1?uOVo.2YFL$\"1^BZ&\\&)el\"FL7$$ \"1s73Uv/tZFL$\"1e#y4)zMh;FL7$$\"1>ZHB-)\\#\\FL$\"1fTKL6Wg;FL7$$\"1b[! 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distinguish the solutions." }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 71 "coltab:=[maroon,red,magenta, yellow,green,blue,violet,brown,gray,black];" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%'coltabG7,%'maroonG%$redG%(magentaG%'yellowG%&greenG% %blueG%'violetG%&brownG%%grayG%&blackG" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "for i from 1 to imax do:" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "theta:=evalf((i-0.5)/imax*Pi/2);" }}{PARA 0 "> " 0 " " {MPLTEXT 1 0 24 "t0:=solve(Y(theta,t),t);" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 71 "P[i]:=plot([X(theta,t),Y(theta,t),t=t0[1]..t0[2]],col or=coltab[i]): od:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "displ ay(seq(P[i],i=1..imax));" }}{PARA 13 "" 1 "" {GLPLOT2D 303 303 303 {PLOTDATA 2 "6)-%'CURVESG6$7U7$$\"182t$)QgcL!#E$\"1\\hk1L38`!#F7$$\"1+ 2kN>*yk(!#;$\"1ZghQB[*>\"F17$$\"1#[bJXcO_\"!#:$\"1J<,Ez0mBF17$$\"1-G5L 0;z@F7$\"1J:))>_^aLF17$$\"1z*HI.Y-$GF7$\"1kS$e`>%=VF17$$\"17\"ok^\"oyU F7$\"1$*HgA$yvR'F17$$\"1bB(R?LYr&F7$\"1%*Hx=$)zn$)F17$$\"1%pg*RH4ArF7$ 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The air density falls off exponentially with increased he ight. The numerical solution of the generalized equation of motion whi ch include a height-dependent air drag coefficient are discussed in N. Giordano's book " }{TEXT 269 21 "Computational Physics" }{TEXT -1 22 " (Prentice-Hall 1997)." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 " " }}}}{MARK "40 0 0" 0 }{VIEWOPTS 1 1 0 3 2 1804 1 1 1 1 } {PAGENUMBERS 0 1 2 33 1 1 }