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What will be its velocity after a random number of seconds?"p  " The computer evaluates your perfor- mance. Answers within 85% of the cor- rect value are accepted. After each question the compknowledge of kinematics. It presents a simple problem: a ball is thrown straight up in the air at some random" h"velocity. You must then answer three questions about the ball's flight:" rj |"1. How high will it go? 2. 30  G' ((GA)A).15280@ :"Not even close."I 300` :"Close enough."j "QQ1 ,:"The correct answer it ";A 6 @ J:8);::"**> PROJECTILE QUIZ <**": T34,2v ^:" This program tests your fundamental o (meters>"; d240 nAV5K x"How long until it returns (seconds>";T 240k T1(2V(1))10x AV10T "What will its velocity be after ";T::"seconds"; 240 Q;" right out of three." Q230 :15)"OK."  PROJECTILE QUIZ":330w:"To continue, touch RETURN":"To quit, enter Q";:Q$:Q$"Q"ĺ(4);"RUN MENU"(:2Q0:V5(35(1))<"A ball is thrown upward at ";V::"meters per second."FPA.05V2 Z"How high will it g  .")75(L$",")*(X1FINISH:X1(J;(" ";H(CURRENTT(VTVT2Z(A: FALLvF4000:PLOP PVT12:CENTER0:WIDTH17ZA$"PRESS THE SPACE BAR TO QUIT.":10000d54,189:55,158n(16384)160130x110: F(LL32P(L$(WRD$,J,1)MZ(L$"@"L$"&"JJ1:LL064(L$"&"):10320Yd(I(L$)mn(I65I90LL0{x((ILL);(SOUND10450(L$" "10450(X12(S(16336)(X(L$" "L$"."L$","10450(FINISH20150(L$" 'WRD$(A$,CURRENT,KCURRENT).'CURRENTKC'LW(WRD$):NLLWd'LW0SPACESSPACES1:10460u'K1(WRD$)'(WRD$,K,1)"@"NLNL1'K(SPACESNLWIDTH10280 (VTVT2(1(SPACE0((SPACESSPACESNL12(VT<(J1LW LOWER CASE PRINTER &'54,0:55,962$'L(A$):.'VTN8'CENTER010120VB'I0aL'K1LkV'II1`'(A$,K,1)"&"(A$,K,1)"@"II1j'Kt'(40I)2~'10130'1'SPACES0'CURRENT1L'KCURRENTL'(A$,K,1)" "ĂK1,1#X1,3:X1,3#X,2%#X6#X2502542D#X,5X,29K#XZ#X271279h#X,5X,50o#X$0,191279,191$1:5$54,0:55,96$"Time: 0""$"Height: 1000",$"Veloc.: 0"6$"Accel.: 0"@$SCREENJ$'786,96 ^"(# SET UP SCREEN 22#SCREEN12H<#SCREEN9030,9040ZF#230,32:9050lP#230,64:9050Z#279,191260,30d#210,30191,191n#260,30191,191x#279,191210,30#140,0279,0#140,4279,4#X2781405#X1,1:X16302,030!1,G9.89Y1(1)26FY1(2)26La@ MUSICAL POKESJ768,173:769,48:770,192:771,136:772,208:773,4:774,198:775,7T776,240:777,8:778,202:779,208:780,246:781,166:782,6:783,76:784,0:785,03:0,0:4430D16299,0&NJ175029X6,J:7,2:768@bJFlYX INITIALIZE xb(768)173(786)967060l8000v(4);"BLOAD MAN,A$7000"(4);"BLOAD CHAR TABLE"(4);"BLOAD HIRES CHAR"232,00233,112 6299,0-SCREENSCREEN1:SCREEN3SCREEN1CSCREEN4320,4330U230,32:4340g230,64:4340v1150,191145,191155,191130,190162,190143,189158,189147,188154,188&149,187152,1870SCREEN4410,4420 :1630ENSCREEN1:SCREEN3SCREEN14TSCREEN4190,4200F^230,32:4210Xh230,64:4210nr1150,Y1(SCREEN){|1150,0143,190157,190146,189153,189147,188152,1886,15:7,1:768SCREEN4280,429016300,0:4300100;" "X6SCREENSCREEN1:SCREEN3SCREEN1LSCREEN4080,4090^230,32:4100p230,64:41001150,Y1(SCREEN)1150,191145,190155,190"149,189151,190,SCREEN4150,4160616300,0:4170@16299,0JSCRE300,0:3180b 16299,0%l J1106v SO(16336)= JD IJ W PLOP bX12sX4007,4008230,32:4010230,6454,0:55,96T1(2SG)5:10:(T1100.5)100;" "10:"0 "10:(T1G100.5)1 SCREEN3040,3050( 230,32:3060: 230,64:3060A 0a X(SCREEN)5,6X(SCREEN),12  X(SCREEN)5,6X(SCREEN)5,6  1X(SCREEN),26  3 & 1150,26 0 145,6146,12 : 155,6154,12 D 145,6155,6 N SCREEN3160,3170 X 16X(SCREEN)5,6X(SCREEN)5,60 H1X(SCREEN),267 R3G \X5,6X,12Y fX5,6X5,6e p1X,26{ zSCREEN2180,2190 16300,0:2200 16299,0 X(SCREEN)X X  RELEASE MAN I12 SCREENSCREEN1:SCREEN3SCREEN1MOVE TO POSITION  SCREEN0% 3; X(1)200:X(2)200M X2401502r SO(16336)(16336)(16336) SCREENSCREEN1:SCREEN3SCREEN1 SCREEN2070,2080 230,32:2090 230,64:2090 *0 4X(SCREEN)5,6X(SCREEN),12 >150,Y1(SCREEN) j1150,Y, tY1(SCREEN)Y= u54,0:55,96G v5:1X w10:T;" "~ x10:((1000S)100.5)100;" " y10:(GT100.5)100;" " z10:G ~SCREEN1160,1170 16300,0:1180 16299,0 EN11020   (4);"RUN MENU"!  DROP MAN . EN0:T0: TT.28O 6,T4:7,2:768c SCREENSCREEN1z SCREEN3SCREEN1 $SCREEN1070,1080 .230,32:1090 8230,64:1090 BS12GT2 LS1000S1000:EN1 VY(1000S).164190 `1[ 7000: INITIALIZE+9000: SCREEN@(2000: POSITIONT23000: RELEASEe<1000: FALLvF4000:PLOP PVT12:CENTER0:WIDTH17ZA$"PRESS THE SPACE BAR TO QUIT.":10000d54,189:55,158n(16384)160130x110:          '" PROJECTILE MOTION"J1.':" ************************"i18'255:A11000:::(7):~1B'"";A$:ACE. LOOK"::"IN YOUR TEXT OR SEE YOUR INSTRUCTOR.":_0v"GOOD LUCK. BYE FOR NOW!":24:100500:D13:(7)::20:"END"0D11500:::(4);"RUN MENU"0?0(#12:2,3737:2,3738:0':100:80'" ************************": 1$2"::7/N"SOLVING--"::7::"Y = 3.67 METERS":::/X"THAT ALL THERE IS TO IT! TYPE 1 TO"::"REVIEW, 2 TO CONTINUE: ";A:A11570/b:6:"THAT'S THE END OF THIS PROGRAM."::Z$;", YOU SHOULD PROBABLY WORK":50l"SOME EXTRA PROBLEMS FOR PRACTIREMEMBER THAT X & Y MOTIONS DO NOT"::"AFFECT EACH OTHER, BUT THE FLIGHT TIME":.0"IS THE SAME. SO, T= 1.3 SECONDS."::10050.::3:"THEREFORE--"::" Y = V(0)Y*T - 1/2*G*T^2":/D:"SUBSTITUTING--"::" Y = 9.192*1.3 - 1/2*9.8*(1.3)^5::"T = 1.3 SECONDS":::10050-:4:"HOW DO WE FIND THE ELEVATION WHERE THE"::"ARROW HITS THE CLIFF? ";:10050::-"HINT: HOW MUCH TIME DOES THE ARROW HAVE"::"TO GO UP AND DOWN?":-"INPUT THE ARROW'S FLIGHT TIME: ";A::Q.&"V(0)X = 12*COS(50) = 7.714 M/S"::" V(0)Y = 12*SIN(50) = 9.192 M/S"::10050,::"SINCE WE KNOW THE RANGE (10 METERS) AND"::"AND THE HORIZONTAL VELOCITY (7.714 M/S),","WE CAN FIND THE FLIGHT TIME--"::" T = RANGE/V(X) = 10/7.714":!-::22:" TIME MARKERS SHOW THE MOTION."K+Y501:7:203Y,11Y2:j+Y03:7:203Y,11Y2:+T11000::24:" TYPE 1 TO SEE AGAIN, 2 TO CONTINUE: ";A:A11620+::2:"HERE'S THE SOLUTION:":::" FIND V(0)X & V(0)Y--":O," 9000:1:X3035:14,36X::7:5,36m*^22:" PRESS TO SHOOT THE ARROW.";A$:T11200:::(7)|*hY50.5*r7:203Y,11Y2:0:203Y,11Y2:*|YO3.5*7:203Y,11Y2:0:203Y,11Y2:*(7):7:29,20)+T11400"HORIZONTAL AND FIRED WITH AN INITIAL":)6"VELOCITY OF 12 M/S TOWARD A VERTICAL"::"CLIFF 10 METERS AWAY. HOW HIGH ABOVE":)@"THE GROUND DOES THE ARROW STRIKE THE"::"CLIFF?"::)J"COPY THE PROBLEM AND THEN WE'LL WATCH.":24:10050%*T:: RANGE = V(X)*T":(" RANGE = 12 * 4"::8::"RANGE = 36 METERS"::::"TYPE 1 TO REVIEW, 2 TO CONTINUE: ";A:A11240(":3:"LET'S DO ONE MORE PROBLEM, THIS TIME"::"IN REVERSE:"::)),"AN ARROW IS AIMED 50 DEGREES ABOVE THE"::-78.4 = 0*T - 1/2*9.8*T^2"::10050:P'"THEN,"::" T^2 = 78.4/4.9":'"AND,"::5::"T = 4 SECONDS":::10050:1550':10:"GREAT WORK, ";Z$;"!"::"KEEP GOING...":22:10050(:5:"SINCE T = 4 SECONDS, WE FIND THE RANGE--"::" - 1/2*G*T^2":::10050y&::"REMEMBER THAT Y = -78.4 METERS, SINCE"::"THE FINAL POSITION IS BELOW THE START."::&" INPUT THE FLIGHT TIME: ";A::A41540&:3:(7):"SORRY, ";Z$;".":::" Y = V(0)Y*T - 1/2*G*T^2":)'"OR,"::" L IS KICKED HORIZONTALLY.":B%" ENTER V(0)Y: ";A::A01460P%"SURE!";%25::"V(0)Y = 0 M/S":::"SINCE THE BALL HAS NO INITIAL VERTICAL"::"VELOCITY. ";:10050&:4:"NOW FIND THE TIME OF FLIGHT--":::"USE: Y = V(0)Y*TR THE SOLUTION. YOU SHOULD BE ABLE"v$x"TO GET THIS ONE ON YOUR OWN.":::"START BY FINDING THE X & Y COMPONENTS":$"OF THE INITIAL VELOCITY."::" ENTER V(0)X: ";A::A121430$"OF COURSE!";%25::"V(0)X = 12 M/S":::"BECAUSE THE BAL5.5/#<9:162Y,11Y2:0:162Y,11Y2:O#F(7):9:26,36:T11200:|#P22:" TIME MARKERS SHOW THE MOTION."#Z9:14,11:Y05:162Y,11Y2::T11000:#d24:" TYPE 1 TO SEE AGAIN, 2 TO CONTINUE: ";A:A11280#$n::2:"NOW FO12 METERS/SECOND?":::"COPY THE PROBLEM. THEN WE'LL LOOK AT IT."::10050y"::9000:3:X515:12,36X::9:13,11" 22:" PRESS TO KICK THE BALL.";A$":T1800::(7)"X02:9:13X,11"(T150:T:0:13X,11:#2Y0"::"HIGH.":22:10050B!:3:"HOW ABOUT ANOTHER PROBLEM?"::!"A MAN KICKS A BALL HORIZONTALLY OFF THE"::"ROOF OF A BUILDING 78.4 METERS TALL.":!"HOW FAR FROM THE BASE OF THE BUILDING"::"DOES IT LAND IF THE INITIAL VELOCITY IS":K""*Y"::" Y = (173.2)^2/(2)(9.8)":X "AND,":9::"Y = 1530.5 METERS"::: "IT'S A BIG CANNON!":::"TYPE 1 TO REVIEW, 2 TO CONTINUE: ";A:A1910 1240!:10:"EXCELLENT WORK, ";Z$;"!"::"IT'S A PRETTY BIG CANNON TO SHOOT THATN HOW":W"HIGH IT GOES.":::"INPUT YOUR MAX. HEIGHT: ";A::A1530A15311230:4:(7):"SORRY. HERE'S THE CORRECT SOLUTION:":"AT MAXIMUM HEIGHT, V(Y) = 0. SO,"::" V(Y)^2 = V(0)Y^2 - 2*G*Y":* " 0^2 = (173.2)^2 - 2*9.8"PROBLEM SOLUTION AGAIN, TYPE 1."::"TYPE 2 TO CONTINUE: ";A:A1910~::"AS A CONTINUATION OF THIS PROBLEM, FIND"::"THE MAXIMUM HEIGHT OF THE SHELL."::10050:3:"HINT: THE FACT THAT THE SHELL MOVES IN"::"THE X DIRECTION HAS NO EFFECT O5 SEC":::YB"* NOW YOU CALCULATE THE RANGE."::" INPUT YOUR ANSWER (NO UNITS):";ApL:5:A35351120~V"GOOD!";`10::"RANGE = V(X)*T = 3535 METERS":::j"THAT'S NOT SO BAD, IS IT?"::::"IF YOU WOULD LIKE TO REVIEW THIS":Gt0) = 0 METERS ";:10050g:3:"USE THIS EQUATION:"::" Y = V(0)Y*T - 1/2*G*T^2"::10050$:"THEN,":" 0 = 173.2*T - 1/2*9.8*T^2":."AND":" -173.2*T = -1/2*9.8*T^2"::10050 8:"FINALLY,"::6::"T = 173.2/4.9 = 35.3S!";79::"V(0)Y = V(0)*SIN(60) = 173.2 M/S":::"NOW USE THE VALUE OF V(0)Y AND THE"::"EQUATIONS OF MOTION FOR GRAVITATIONAL":"ACCELERATION TO FIND THE TIME OF FLIGHT."::"REMEMBER THAT THE SHELL HITS THE GROUND,":"SO: Y = Y(SWER FOR V(0)X: ";A::A100970/"GOOD!";f10::"V(0)X = V(0)*COS(60) = 100 M/S":::10050:3:"REMEMBER THAT THIS IS THE VALUE OF THE"::"CONSTANT HORIZONTAL VELOCITY."::"INPUT YOUR VALUE FOR V(0)Y: ";A::A173.21010"YERED WITH A MUZZLE":k"VELOCITY OF 200 M/S AT AN ANGLE OF"::"60 DEGREES ABOVE HORIZONTAL. FIND THE":"RANGE OF THE SHELL. ";:10050::"COPY THE PROBLEM. THEN TRY TO FIND THE"::"X & Y COMPONENTS OF V(0).":!:"INPUT YOUR AN(0) = 0, Y(0) = 0.":jz"> THE STARTING TIME IS T(0) = 0."::"> THE ANGLE (THETA) IS MEASURED FROM ":" THE X-AXIS.":::"COPY THESE CONVENTIONS IF YOU NEED TO."::"THEN, ";:10050:3:"NOW HERE'S A PROBLEM:":::"A CANNON SHELL IS FICE: ";:10050q\:2:"THE CONVENTIONS WE USE ARE THE SAME AS"::"THOSE ESTABLISHED IN PREVIOUS PROGRAMS:"::f"> UP AND RIGHT ARE POSITIVE."::"> DOWN AND LEFT ARE NEGATIVE.":p"> THE STARTING POSITION (EVEN IF ABOVE"::" THE GROUND IS X V(0)X!"::" V(0)Y = V(0)*SIN(THETA)"}>" NOTE: V(Y) WILL CHANGE!"::" 4. APPLY EQUATIONS OF MOTION IN THE"H" Y DIRECTION TO FIND THE TIME OF":" FLIGHT (T)": R" 5. FIND RANGE = V(X) * T"::"COPY THE METHOD FOR REFEREN ";:10050B:2:"METHOD:"::" 1. ESTABLISH CONVENTIONS": " 2. DETERMINE KNOWN & UNKNOWN VALUES"::" 3. FIND THE X & Y COMPONENTS OF THE"*" INITIAL VELOCITY--"::" V(0)X = V(0)*COS(THETA)",4" NOTE: V(X) = CONSTANT"THE PROJECTILE. YOU ARE TO SOLVE FOR THE":"RANGE (MAX. HORIZONTAL DISTANCE)."::"OF COURSE, THE PROBLEM CAN BE CHANGED"::"AROUND (GIVEN RANGE AND ANGLE, FIND THE":  "INITIAL VELOCITY). BUT, LET'S START WITH":"THE EASY ONE! ";Z$;"?"::"TYPE 1 TO SEE AGAIN, 2 TO CONTINUE: ";A:A1370:3:"NOW LET'S LOOK AT A METHOD FOR SOLVING"::"PROJECTILE PROBLEMS. ";:10050::"IN THE BASIC PROBLEM, YOU ARE GIVEN THE"::"INITIAL VELOCITY AND ANGLE (THETA) OF":Vz #???       Ѡ ήԠ Ѡ+Ϡà"Ҡ® ҠՠҠ"͠+N AND BY THE"::"ACCELERATED MOTION IN THE Y DIRECTION."::"THEREFORE, THE TIME IT TAKES THE PRO-"::"JECTILE TO GO UP AND DOWN IS EXACTLY":"THE AMOUNT OF TIME THE PROJECTILE HAS"::"TO MOVE SIDEWAYS."::>"DO YOU WANT TO WATCH AGAIN,1000:TC11:10,36203Y:7:21,3611Y2:1:203Y,11Y2:24:" TYPE 1 TO SEE AGAIN, 2 TO CONTINUE: ";A:A1620::3:"YOU SHOULD SEE THAT THE PROJECTILE"::"POSITION IS DETERMINED BY THE CONSTANT":="VELOCITY IN THE X DIRECTIO>b24:" TYPE 1 TO SEE AGAIN, 2 TO CONTINUE: ";A:A1370l::9000:1:5,36:22:" HERE IT IS IN SLOW MOTION.":T11500:vY501:T11000:T11:10,36203Y:7:4,2011Y2:1:203Y,11Y2:7:21,3611Y15:T1000:Y50.5?13:203Y,11Y2:0:203Y,11Y2:M&YO5.5x013:203Y,11Y2:0:203Y,11Y2::13:35,36DT11000::22:" TIME MARKERS SHOW THE MOTION."NY501:13:203Y,11Y2:XY05:13:203Y,11Y2:.5$7:4,11Y2:0:4,11Y2:2Y05.5[7:4,11Y2:0:4,11Y2::7:4,36T11000::22:" TIME MARKERS SHOW THE ACCELERATION."7:Y05:4,11Y2::24:10050:22:" NOW COMBINE THE MOTIONS."13:5,36:T111000: Y010.1B 11:53Y,10:0:53Y,10::11:35,10 T11000::22:" TIME MARKERS SHOW CONSTANT VELOCITY. " 11:Y010:53Y,10::24:10050 :22:" HERE'S A VERTICAL ACCELERATION." 7:4,36:T11000:Y50ILE MOTION AS TWO":[ ^"SEPERATE MOTIONS WHICH LAST FOR THE"::"SAME AMOUNT OF TIME.": h:"LET'S SEE HOW THE MOTIONS COMBINE."::"WATCH CAREFULLY, ";Z$;"!":::10050 r::9000:22:" HERE'S A CONSTANT HORIZONTAL VELOCITY." |11:5,10:T1N DUE TO"+ ," GRAVITY"::6 610050 @:3:"THESE TWO MOTIONS OCCUR AT THE SAME"::"TIME, AND THE RESULT IS PARABOLIC": J"PROJECTILE MOTION. ";:10050:: T"THE TRICK IN SOLVING PROBLEMS IS TO"::"CONSIDER THE PROJECT:[ "PROJECTILE MOTION CAN BE CONSIDERED AS"::"TWO SIMULTANEOUS MOTIONS WHICH ARE":r :"INDEPENDENT": 12:13:"EXECPT FOR THE TIME FACTOR.":::"THE MOTIONS ARE:": "" 1. HORIZONTAL - CONSTANT VELOCITY"::" 2. VERTICAL - ACCELERATIOFFICULTY WITH":_ "THEM--SO PAY CLOSE ATTENTION.":::"YOU SHOULD HAVE ALREADY WATCHED:": 9)"KINEMATICS AND GRAVITY"::"WHEN YOU'RE READY, ";:10050 :3:"THE BASIC IDEA IS SIMPLE, BUT YOU MUST"::"UNDERSTAND IT WELL TO SOLVE PROBLEMS.":)d- PROJECTILE MOTION%n10000[:3:"HELLO!"::"PLEASE TELL ME YOUR NAME: ";Z$8:"THANK YOU, ";Z$;". THIS PROGRAM"::"IS DEVOTED TO THE SOLUTION OF PROJECTILE" "PROBLEMS. ALTHOUGH THEY'RE REALLY NOT "::"TOO HARD, STUDENTS HAVE DI             4D6,Z:7,2:768D:)DVTVT2/EA䠠䮠󬠨䠨᠍򬠠堍뮍堠Է¹Ӎ䠍砠󬠠堠堠 L):15050::1:(hB HIRES INPUT 6rB16368,0A|B1:VTMB"===>"VBHT4`BG$""pBA(16384)BVT:HT1:" "BVT:HT1:(95)BA12717060B16368,0BA14117600BA15517580BA14717160BSOUNDSOUND1:SX,Y #X#T$:^$''$h$:hr$ 173,48,192,136,208,4,198,7,240,8,202,208,246,166,6,76,0,3,96~: PRESS ANY KEY :16368,0:CU$"!/-\":24:"! press any key":C1:L4:24:1:(CU$,C,1) :KEY(16384):KEY128CC1L(C16336)d6240j*(# INITIALIZE 52#16384K<#(24576)729050hF#(4);"BLOAD HIRES CHAR"P#(4);"BLOAD CHAR TABLE"Z#VT1:SOUND0:CENTER0:WIDTH40d#G9.8n#x#XS3:YS3#TR$(0)"OFF":TR$(1)"ON"#X768786#Y# XXSD5Y183HYS+X2796230:TR16180A0LX1,Y1[H06230iY06220p3yX,Y6210$Y06210.X,Y8H06230BX1X:Y1YL6040V16368,0`(16384)1606250bSS1:S5S0:J(5190(Y1831010&2X,YX,Y5-<Y@FDOT(3)05260MPDT2VHZZX5DXSndX5X2795260nY1831010xX,YX,Y5Yp THROW BALL zT0:X10:Y1035,183TT.05H12GTTVVTDVHTTHETAAN3.141592653180?VHV0(THETA):VVV0(THETA)NT(0VV)GcS12GTTVVTvDOT(1)05120Y183SYSY0Y18351203X526510X,YX5,YXDOT(2)05190 DTVHX5DXSX5X279183279,183X0(274XS))Y1187@X5(X5)Y1189YX10(X10)Y1191sXXS5,183XXS5,Y1zXY0(183YS)X13Y5(Y5)X11"Y10(Y10)X10,5,183YYSX1,183YYS6Y@ dotted lines stance."/6 "3. point where it hits ground.":@ J13GJ J3:35bT DOT(J)1ĺ"Yes":2920m^ "No "th Jr VT10:17000| G$"G"2010 J(G$) DOT(J)DOT(J)12(DOT(J)1) 2880  SET UP GRID 3 5,05, :"the graph." VT11( 170005 XS(G$)? 2010T TRTR12(TR1)^ 2010g :1 "Where would you like the dotted lines?" "(press 'G' to Go back to menu)"" :"1. greatest height reached.", "2. midpoint of total di(G$)*x XS0XS6XS(XS)Đ:VT1:2540: XS62790C :1p "Please enter a number to indicate the" :"maximum distance on the graph. When you" "divide 274 by this number, you will have" "the number of meters that will fit on"re for the": :"maximum distance that will fit on the"g :"screen. Which one would you like??"m } "1. 274m"( "2. 137m"2 "3. 92m"< "4. 68m"F "5. 54m"P "6. other"Z :"enter your choice"d VT16:17000n XSicate the": :"maximum height on the graph. When you"l :"divide 183 by this number, you will have" "the number of meters that will fit on" :"the graph." VT11 17000 YS(G$) 2010 :1  "The following choices a?"  "1. 183m"'$ "2. 91m"6. "3. 61m"E8 "4. 45m"TB "5. 36m"eL "6. other"V :"enter your choice"` VT16:17000j YS(G$)t YS0YS6YS(YS)Đ:VT1:2280~ YS62530 :1  "Please enter a number to indf inclination." 17000' AN(G$)4 Y1823f Y:" "m Yw 2020 :1 "The following are choices for the" :"maximum height that will fit on the" :"graph. Which one would you like?pCH0CH92010A zCH2180,2230,2280,2540,2800,2820,2980,100O 18:VT20r "Enter new initial velocity."} 17000 V0(G$) Y1823 Y:" " Y 2020 18:VT20 "Enter new angle o:"4. max X (distance) ";(274XS);" "X :"5. Trace mode ";TR$(TR);" "` :x *"6. Dotted lines" 4:"7. Graph projectile" >:"8. End and return to MAIN MENU" H:"Enter your choice." RVT20 \17000 fCH(G$) "<<"$"$"*$"""$&"$"* > <>LL@0   @0 @LL@@@0 @ @0 @@@$&"&"<*0  00L@00 """ 0   00@@00 <$ """""""<< "2""LL@0   @0 @LL@@@0 @ @0 @@@$""2||p@|@|p||p*0L 0 @@00 $$ @ """""* """  "@@0|  pp@p@@LL0 @@ 0LL"$$" "||p@|@|p||p>0L 0 @@00>> @a"> >>88 @@0|  pp@p@@LL0 @@ 0LL>          e computer's high resolution". :"screen."9 15000? 8W  GET INFORMATION ] d 1u 54,0:55,96 "1. Initial velocity ";V0;"m/sec " :"2. Angle of inclin. ";AN;(2);" " :"3. max Y (height) ";(183YS);" "+ n experiment with projectile motion"S :"with the use of this last part of the"e :"program." ::"It allows you to enter an initial" :"velocity and angle of inclination of a" :"projectile and see the path traced out" :"on th0  900010002000)(40003-5000=26000E<30Yd54,189:55,158rn::(4);"RUN MENU" FINAL WORDS 54,0:55,96:1"This is the end of the formal":"demonstration. If you would like, you"$ :"ca       4501,-()$% !<<884400,,(($$ ` 3 $[0 Ʉɑ`LB#( ??;;7733//++''##>?:;6723./*+&'"#>>::6622..**&&""=>9:5612-.)*%&!"==995511--))%%!!<=89.(**(&&(""=(>9(:5(61(2-(.)(*%(&!("(==(99(55(11(--())(%%(!!<(=8(94(50(1,(-(()$(% (!(<<(88(44(00(,,(((($$( ?(<;(87(43(0/(,+(('($4P50P1,P-(P)$P% P!P<(?:(;6(72(3.(/*(+&('"(#(>>(::(66(22(.ιP??P;;P77P33P//P++P''P##>P?:P;6P72P3.P/*P+&P'"P#P>>P::P66P22P..P**P&&P""=P>9P:5P61P2-P.)P*%P&!P"P==P99P55P11P--P))P%%P!!"&*.26:>#'+/37;?#'+/37;? $(,048< $(,048"&*.26:>#'+/37;?#'+/37;? $(,048< $(,048"&*.26:>#'+/37;?#'+/37;?  Q`((((((((((((((((((((((((((((((((PPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPО PRW` @ @ @ @ii``󠠠D$;"NOMON C,I,O"XD$;"BLOAD ";F$;",A$3B8"3:"APPLE ][ EDITOR-ASSEMBL %  Q  Q  i0 i ` PRW  ` ʽ  %ȑ8ؠ    Л`l   I` 8'ȱ     ,2:768(CX:Y(CVTVT2#(C040<(CSOUND17370G(CY13V(CX4014i(C6,X:7,2:768s(CX:Y(CVTVT2(C(C6,X:7,2:768(CX:Y(CVTVT2(C "'&CG$""'0C17040$':CLL13'DCG$(G$,L)E'NCVT:L5:" "P'XC17040c'bCG$G$(A128)o'lCL(G$)'vCVT:L4:(A)'{CSOUND17310'CX2012'C6,X:7,2:768'CX'C17040'CSOUND17370'CY13'CX4014(C6,X:76384)&BA12717040$&B16368,05&BA14117320G&BA13617250Y&BSOUND17120g&BX1205z&B6,X:7,2:768&B&BL(G$)&BL117210&BL117180&BSOUND17180&BY110&BX1256&C6,X:7,3:768&C:'CVT:5:" NY KEY %:16368,0*%:CU$"!/-\":VT24J%:A$"! PRESS ANY KEY":10000^%:VTVT2:C1:L4u%:VT:1:(CU$,C,1)%:KEY(16384):KEY128CC1L(CL):15050%::VT1:%hB HIRES INPUT %mB16368,0%rB1:VT%|B"===>"%BG$""&BA(1LL0$x((ILL);$$(SOUND104506$(L$" "10450A$(X12Q$(S(16336)X$(X{$(L$" "L$"."L$","10450$(FINISH20150(L$" .")75(L$",")$(X1FINISH:X$(J$(" ";$(CURRENT$(VTVT2$(%: PRESS AK1(WRD$))#'(WRD$,K,1)"@"NLNL10#'KL#(SPACESNLWIDTH10280X# (VTVT2_#(1k#(SPACE0#((SPACESSPACESNL1#2(VT#<(J1LW#F(LL32#P(L$(WRD$,J,1)#Z(L$"@"L$"&"JJ1:LL064(L$"&"):10320#d(I(L$)$n(I65I901)"`'(A$,K,1)"&"(A$,K,1)"@"II10"j'K>"t'(40I)2I"~'10130P"'1]"'SPACES0n"'CURRENT1L"'KCURRENTL"'(A$,K,1)" "ĂK"'WRD$(A$,CURRENT,KCURRENT)"'CURRENTK"'LW(WRD$):NLLW"'LW0SPACESSPACES1:10460 #'#16384!<#(24576)729050:!F#(4);"BLOAD HIRES CHAR"W!P#(4)"BLOAD CHAR TABLE"z!Z#VT1:SOUND0:CENTER0:WIDTH40!d#G9.8!n#!x#!' LOWER CASE PRINTER !'54,0:55,96!$'L(A$)!.'VT!8'CENTER010120!B'I0!L'K1L"V'II0:VV(VV10.5)10 !TT.054 !H12GTTVVT? !DVHTK !X3D5Y !Y183H3h !TR18720o !0z !X1,Y1 !H08760 !3 !X,Y !8740 "H08760 "H08760 "X,Y $"X1X:Y1Y ."8590 8" (# INITIALIZE !2VH:D(D10.5)10l X5D30v Y1832010? X,YX,Y5F YU TR0:8500d TR1:8500j }4! THROW BALL >!T0:X10:Y10H!3R!5,183\!V030f!AN45p!ANAN3.141592653180z!VHV0(AN):VVV0(AN) !VH(VH10.5)1X12Y5(Y5)X111Y10(Y10)X10K5,183Y3X1,183Y3RYXy CONTROL THROWING OF BALL  TR0:8500 T(0VV)G:T(T10.5)10& S12GTTVVT0 Y183S3: 3D X526510N X,YX5,YX Xb DTON OF THIS MOTION. @NOTICE HOW IT CONFORMS TO OUR INFORMATION.":10000R15000Xl@ SET UP GRID sJ3T5,05,183279,183^X091hY1187rX5(X5)Y1189|X10(X10)Y1191X35,183X35,Y1XY060A$"S = V T":10000"VTVT1:A$" H ":10000GCENTER0RSVHTA$"WE FIND THAT AT "(T)" SECONDS, THE BALL HAS TRAVELLED "(S)" METERS. @THIS IS THE ANSWER TO THE SECOND PART.":10000VTVT1GA$"@HERE IS A DEMONSTRATI@SINCE IT TAKES "(T)" SECONDS TO GO UP, IT MUST ALSO TAKE "(T)" SECONDS TO GO DOWN, SO IT WILL HIT THE GROUND AT "(T2)" SECONDS.":1000015000X SECOND PART OF PROBLEM bTT2lA$"@USING THE EQUATION:":10000vCENTER1BT H ,L h0` 2 ":100005ZA$" V ":10000BdCENTER0WnS12GTTVVTkxS(S10.5)10A$"AND WE GET HEIGHT = "(S):10000A$"@THIS IS THE ANSWER TO THE FIRST PART.":10000VTVT1A$"@THE SECOND PART IS EASY NOW. 10000:CENTER0 CENTER0tA$"@THIS MEANS THAT AT "(T)" SECONDS, THE BALL IS AT ITS HIGHEST POINT.":1000015000(A$"@WE NOW INSERT THIS INTO THE EQUATION:":100002CENTER1<A$"S = 1/2 GT + V T":10000FVTVT3PA$" 10000CENTER1/A$"V = GT + V ":10000;VTVT1WA$" O":10000fT(0VV)GzT(T10.5)10CENTER0A$"@WHEN YOU INSERT THE DATA INTO THE EQUATION AND SOLVE FOR 'T', YOU GET:":10000CENTER1A$"T = "(T): "(TH)(2)" = "(VV):10000,tVTVT1f~A$" V"(" ",(A$)5):10000sCENTER0~15000A$"@TO FIND HOW HIGH THE BALL WILL RISE, WE USE THE FACT THAT THE VELOCITY IS ZERO AT THE HIGHEST POINT, AND THE EQUATION:":0VHV0(AN)!$VVV0(AN)7.VH(VH10.5)10M8VV(VV10.5)10ZBCENTER1LA$"V = "(V0)" X @C@O@S "(TH)(2)" = "(VH):10000VVTVT1`A$" H"(" ",(A$)5):10000 jA$"V = "(V0)" X @S@I@N10000VTVT1.A$" H O ":10000OA$"V = V @S@I@N @A":10000[VTVT1xA$" V O ":10000CENTER0A$"@WHEN WE PUT THE GIVEN INFORMATION INTO THESE FORMULAS, WE GET...":10000V030:TH45ANTH3.14159265318LL GET, AND HOW FAR AWAY WILL IT LAND.":10000915000A$"@FIRST, WE HAVE TO FIGURE OUT THE HORIZONTAL AND VERTICAL COMPONENTS FROM THE GIVEN DATA. @THE FORMULAS THAT HAVE TO BE USED TO DO THIS ARE:":10000CENTER1A$"V = V @C@O@S @A":RE IS A BRIEF DEMONSTRATION PROBLEM TO ILLUSTRATE HOW TO SOLVE PROJECTILE MOTION PROBLEMS.....":10000rVTVT1.A$"@A BALL IS FIRED FROM A CANNON WITH AN INITIAL VELOCITY OF 30 M/SEC AND AN ANGLE OF INCLINATION OF 45"(2)". @HOW HIGH WILL THE BA0 d5230nVTVT1xA$"@THIS IS WHY YOU WILL HAVE TO FIND THE TIME THAT A PARTICLE STOPS BY USING THE VERTICAL COMPONENT, AND THEN USING THAT TIME IN THE HORIZONTAL COMPONENT FORMULA.":1000015000p HORIZONTAL PROBLEM fzA$"@HE((G$)47(G$)5852102A$"@I CAN'T FIGURE OUT WHAT YOUR ANSWER IS, BUT THE CORRECT ANSWER IS THAT IT WILL NEVER STOP!":10000<5230FA$"@THAT'S CORRECT!":10000P5230ZA$"@THAT'S NOT CORRECT. @THE SPACESHIP WILL NEVER STOP.":10001000017000"G$"H"5100MA$"@NO HINTS ON THIS PROBLEM!":10000X17000c:VT1(G$,8)"IT WON'T"5190(G$,11)"IT WILL NOT"5190 (G$,13)"IT WILL NEVER"5190(G$,5)"NEVER"5190(G$,10)"AT NO TIME"5190T TA$"@NOW LET'S EXAMINE THE HORIZONTAL COMPONENT OF PROJECTILE MOTION.":10000iCENTER1:VTVT1A$"PROBLEM 2":10000CENTER0A$"@A SPACESHIP IS TRAVELLING AT 1000 M/SEC IN A FRICTIONLESS ENVIRONMENT. @AT WHAT TIME WILL IT STOP?":V0)GS12GTTV0T.S(S10.5)10C(SAN).64410oA$"@SORRY, THAT'S NOT CORRECT.":10000&A$"@THE CORRECT ANSWER IS "(S)" METERS.":1000004420:A$"@THAT IS CORRECT!!":10000D15000N HORIZONTAL COMPONEN+ V T":10000 VTVT39 A$" 2 ":10000Y A$" O ":10000f CENTER0 A$"@HOW HIGH (IN METERS) DOES THE BALL GET WITH AN INITIAL VELOCITY OF "(V0)" M/SEC?":10000 17000 :VT1 AN(G$) V0T(Vɺ  HELLO (CHEMISTRY SERIES),D$(4)FD$;"BLOAD POUR SUB."b(D$;"BLOAD TPIC,A$4000"l22560<D$;"BLOAD MR"F1013,76:1014,0:1015,3PA$:ZD$;"RUN MENU"NU"AZERO AT THE HIGHEST POINT.":10000^ hA$"@USE THE EQUATION: V=GT+VO AND SOLVE FOR T.":10000 rA$"@THIS IS THE TIME AT WHICH THE BALL REACHES ITS HIGHEST POINT.":10000 |A$"@NOW SUBSTITUTE INTO THE EQUATION:":10000 CENTER1 A$"S=1/2 GT T BEGINS TO FALL BACK TOWARDS THE @EARTH?":10000= ,VTVT1z 6A$"@ENTER YOUR ANSWER OR '@H' IF YOU NEED HELP.":10000 @17000 J(G$,1)"H"4310 T:VT1" ^A$"@TO COMPUTE THE ANSWER TO THIS PROBLEM, FIRST USE THE FACT THAT THE VELOCITY IS VERY SIMPLE PROBLEM.":10000' 15000@ V0(((8)10)1)10M CENTER1h A$"@PROBLEM 1":10000u CENTER01 "A$"@A TENNIS BALL IS THROWN DIRECTLY UPWARDS WITH AN INITIAL VELOCITY OF "(V0)" M/SEC. @HOW HIGH (IN METERS) WILL IT GET BEFORE IE SAME TIME TO GO UP AS IT DID TO GO DOWN.":10000 A$"2. @THE VELOCITY CHANGED FROM POSITIVE TO NEGATIVE AT THE HIGHEST POINT (IT WAS ACTUALLY ZERO AT THIS POINT)." 10000:VTVT1 A$"@NOW LET'S USE THESE TWO FACTS TO COMPUTE THE ANSWER TO A $  900040005000)(6000327000=<8000GF8200[P54,189:55,158rZ(4);"RUN PRO.VI" EXPLANATION A$"@SOME THINGS YOU SHOULD HAVE NOTICED IN THIS EXAMPLE ARE:":10000VTVT12 A$"1. @THE BALL TOOK TH          `H XN7<6NWRU@ TPhhHNɍ-Ɍ+*+*++8 $%ll%J) +%j )* * (je$*h-HJJJ'hH*&'*&'*&')&&OH**hQ**O+i+$$! $%%#"%Nh``H XN7<6NWRU@ TPhhHNɍ-Ɍ+*+*++8 $%ll%J) +%j )* * (je$*h-HJJJ'hH*&'*&'*&')&&OH**hQ**O+i+$$! $%%#"%Nh`RUN SC" rD$;"RUN MS") |D$;"RUN TR"APQ"A@D$;"RUN PL"& JD$;"RUN PM"7 TD$;"RUN PQ"AREADER"T @D$;"RUN GFL"e JD$;"RUN PL"v TD$;"RUN PM" ^D$;"RUN PQ" hD$;"RUN FF"L rD$;"RUN MS" |D$;"RUN TR";MG" JD$;"RUN CF" TD$;"RUN AA" ^D$;"RUN TT" hD$;"8:T)"(1) INSTRUCTIONS"; x:T)"(2) PROJECTILE MOTION"[ :T)"(3) PROJECTILE QUIZ"w :T2)"< TOUCH KEY #" 14:T);">";:A$:A$ A(A$) A1A610 A310,320,330,340,350,360 6D$;"RUN READER" @D$;"RUN PM" JD$;"RUN  PROJECTILES MENU134,0:D$(4)::255K80: JUMP TO PROGRAM]( SUBROUTINES2:8)"TOUCH KEY TO CONTINUE =>";:A$::F ** MAIN MENU **P:8);::"*>>> PROJECTILES <<<*":Z:8)"(c) 1989 by Adrian Vance"dT9 n     (; !) )###'  . #<$)-2(L$",")(X1FINISH:X"(J,(" ";9(CURRENTE(VTVT2K($(WRD$,J,1)>Z(L$"@"L$"&"JJ1:LL064(L$"&"):10320Jd(I(L$)^n(I65I90LL0lx((ILL);~(SOUND10450(L$" "10450(X12(S(16336)(X(L$" "L$"."L$","10450(FINISH20150(L$" .")75URRENT,KCURRENT)'CURRENTK4'LW(WRD$):NLLWU'LW0SPACESSPACES1:10460f'K1(WRD$)'(WRD$,K,1)"@"NLNL1'K(SPACESNLWIDTH10280 (VTVT2(1(SPACE0((SPACESSPACESNL12(VT<(J1LWF(LL32 P(LINTER '54,0:55,96#$'L(A$)+.'VT?8'CENTER010120GB'I0RL'K1L\V'II1`'(A$,K,1)"&"(A$,K,1)"@"II1j'Kt'(40I)2~'10130'1'SPACES0'CURRENT1L'KCURRENTL'(A$,K,1)" "ĂK'WRD$(A$,CX1,3#X,2#X'#X25025425#X,5X,29<#XK#X271279Y#X,5X,50`#Xs$0,191279,191}$1:5$54,0:55,96$"Time: 0""$"Height: 1000",$"Veloc.: 0"6$"Accel.: 0"@$SCREENJ$' LOWER CASE PR(# SET UP SCREEN #2#SCREEN129<#SCREEN9030,9040KF#230,32:9050]P#230,64:9050qZ#279,191260,30d#210,30191,191n#260,30191,191x#279,191210,30#140,0279,0#140,4279,4#X2781405#X1,1:X1,1#X1,3:3 01G9.8*Y1(1)267Y1(2)26=R@ MUSICAL POKESJ768,173:769,48:770,192:771,136:772,208:773,4:774,198:775,7T776,240:777,8:778,202:779,208:780,246:781,166:782,6:783,76:784,0:785,03:786,96^16299,0NJ17502*X6,J:7,2:7681bJ7lJX INITIALIZE ib(768)173(786)967060sl8000v(4);"BLOAD MAN,A$7000"(4);"BLOAD CHAR TABLE"(4);"BLOAD HIRES CHAR"232,00233,11216302,0ENSCREEN1:SCREEN3SCREEN14SCREEN4320,4330F230,32:4340X230,64:4340g1150,191|145,191155,191130,190162,190143,189158,189147,188154,188&149,187152,1870SCREEN4410,4420:16300,0:4430DREEN3SCREEN1%TSCREEN4190,42007^230,32:4210Ih230,64:4210_r1150,Y1(SCREEN)l|1150,0143,190157,190146,189153,189147,188152,1886,15:7,1:768SCREEN4280,429016300,0:430016299,0SCRE'SCREENSCREEN1:SCREEN3SCREEN1=SCREEN4080,4090O230,32:4100a230,64:4100w1150,Y1(SCREEN)1150,191145,190155,190"149,189151,190,SCREEN4150,4160616300,0:4170@16299,0JSCREENSCREEN1:SC 16299,0l J110'v SO(16336). J5 I; H PLOP SX12dX4007,4008v230,32:4010230,6454,0:55,96T1(2SG)5:10:(T1100.5)100;" "10:"0 "10:(T1G100.5)100;" "X40,3050 230,32:3060+ 230,64:30602 0R X(SCREEN)5,6X(SCREEN),12t  X(SCREEN)5,6X(SCREEN)5,6  1X(SCREEN),26  3 & 1150,26 0 145,6146,12 : 155,6154,12 D 145,6155,6 N SCREEN3160,3170 X 16300,0:3180 bX(SCREEN)5,6! H1X(SCREEN),26( R38 \X5,6X,12J fX5,6X5,6V p1X,26l zSCREEN2180,2190 16300,0:2200 16299,0 X(SCREEN)X X  RELEASE MAN I12 SCREENSCREEN1:SCREEN3SCREEN1 SCREEN30N  SCREEN0 3, X(1)200:X(2)200> X2401502c SO(16336)(16336)(16336) SCREENSCREEN1:SCREEN3SCREEN1 SCREEN2070,2080 230,32:2090 230,64:2090 *0 4X(SCREEN)5,6X(SCREEN),12 >X(SCREEN)5,6 j1150,Y tY1(SCREEN)Y. u54,0:55,968 v5:1I w10:T;" "o x10:((1000S)100.5)100;" " y10:(GT100.5)100;" " z10:G ~SCREEN1160,1170 16300,0:1180 16299,0 EN11020   MOVE TO POSITIO  DROP MAN  EN0:T0+ TT.28@ 6,T4:7,2:768T SCREENSCREEN1k SCREEN3SCREEN1 $SCREEN1070,1080 .230,32:1090 8230,64:1090 BS12GT2 LS1000S1000:EN1 VY(1000S).164190 `1150,Y1(SCREEN)L 7000: INITIALIZE+9000: SCREEN@(2000: POSITIONT23000: RELEASEe<1000: FALLvF4000:PLOP PVT12:CENTER0:WIDTH17ZA$"PRESS THE SPACE BAR TO QUIT.":10000d54,189:55,158n(16384)160130x110:        <"\:FBF:<<@@\bBb\\bb\@<:FBBB8 0 "" "8.TTTT>DDDD8DDD8:FF:\bb\@@:F|<@>>H0BBBb\BBB$DDTTlB$$BBBb\@<~ ~880(8DRj2x$B~BBB>DD
DDDDD>~~~BB>BB>"B$$$$$~$~$$x8P<F&db  R"\   T8|8T| ~@ BB<~B > 60HH0>>>***>"6*""">$ 88>>>>*>><> ?:{>     <"\:FBF:<<@@\bBb\\bb\@<:FBBB8 0 "" "8.TTTT>DDDD8DDD8:FF:\bb\@@:F|<@>>H0BBBb\BBB$DDTTlB$$BBBb\@<~ ~880(8DRj2x$B~BBB>DD
DDDDD>~~~BB>BB>"B$$$$$~$~$$x8P<F&db  R"\   T8|8T| ~@ BB<~B > 60HH0>>>***>"6*""">$ 88>>>>*>><> ?:{>  L <"<"""<< <"""<"><$""< """"  "" 6***""""""""""<""< :< $"""2,"""**6""">>8  80,"*:<>""""""""""""">>><2"<""">""" ""  ">"6**"""""&*2""""""""""""*,"" "" ">""""""""""""**6"""">>>> >00000>"(~ < (&20 *, **>> """ " >> ">> """>""""""< >> "G$)vCVT:L5:(A)({CSOUND173107CX2012JC6,X:7,2:768QCX\C17040nCSOUND17370yCY13CX4014C6,X:7,2:768CX:YCVTVT2C/6,X:7,2:768B BL(G$)/BL117210>BL117180PBSOUND17180\BY110jBX1256}C6,X:7,3:768C:CVT:5:" "&CG$""0C17040:CLL1DCG$(G$,L)NCVT:L6:" "XC17040bCG$G$(A128)lCL(:VT:1:(CU$,C,1)G:KEY(16384):KEY128CC1L(CL):15050T::VT1:hhB HIRES INPUT srB1:VT|B"===>"BG$""BA(16384)BA12717040B16368,0BA14117320BA13617250BSOUND17120BX1205BL$" "L$"."L$","10450P(FINISH20150(L$" .")75(L$",")c(X1FINISH:Xj(Jt(" ";(CURRENT(VTVT2(: PRESS ANY KEY :16368,0:CU$"!/-\":VT24:A$"! PRESS ANY KEY":10000:VTVT2:C1:L41 (SPACE0$((SPACESSPACESNL1,2(VT8<(J1LWBF(LL32UP(L$(WRD$,J,1)Z(L$"@"L$"&"JJ1:LL064(L$"&"):10320d(I(L$)n(I65I90LL0x((ILL);(SOUND10450(L$" "10450(X12(S(16336)(X('CURRENT1L!'KCURRENTL8'(A$,K,1)" "ĂKY'WRD$(A$,CURRENT,KCURRENT)g'CURRENTK|'LW(WRD$):NLLW'LW0SPACESSPACES1:10460'K1(WRD$)'(WRD$,K,1)"@"NLNL1'K(SPACESNLWIDTH10280 (VTVT2(l%X:HT:A$&v%6,X230:7,2:768-%X3%N' LOWER CASE PRINTER _'54,0:55,96k$'L(A$)s.'VT8'CENTER010120B'I0L'K1LV'II1`'(A$,K,1)"&"(A$,K,1)"@"II1j'Kt'(40I)2~'10130'1'SPACES0$"WRITTEN BY":S5:9500:J$A$"LARRY SELTZER":S7:9500KT$J11000:JQ^$l% WORDS AT BEGINNING x&%L(A$)0%HT(40L)2:%X2301D%X1X5:X1249570N%X1S9610X%X1:" ";b%XS9590050::VT1:#Y23114#6,Y:7,2:768@#XYABM#X1X1W#Y:X_#L$k#Y1:X1t#" "#X1X:Y1Y#Y#1:15#(WRD$,T)#T$T34171$2:T$"MOTION ""$J150:,$6,T:7,2:7686$@$A75(L$",")(X1FINISH:X&(J0(" ";=(CURRENTI(VTVT2O(e: PRESS ANY KEY s:16368,0:CU$"!/-\":VT24:A$"! PRESS ANY KEY":10000:VTVT2:C1:L4:VT:1:(CU$,C,1):KEY(16384):KEY128CC1L(CL):15P(L$(WRD$,J,1)BZ(L$"@"L$"&"JJ1:LL064(L$"&"):10320Nd(I(L$)bn(I65I90LL0px((ILL);(SOUND10450(L$" "10450(X12(S(16336)(X(L$" "L$"."L$","10450 (FINISH20150(L$" .")A$,CURRENT,KCURRENT)#'CURRENTK8'LW(WRD$):NLLWY'LW0SPACESSPACES1:10460j'K1(WRD$)'(WRD$,K,1)"@"NLNL1'K(SPACESNLWIDTH10280 (VTVT2(1(SPACE0((SPACESSPACESNL12(VT<(J1LWF(LL32E PRINTER '54,0:55,96'$'L(A$)/.'VTC8'CENTER010120KB'I0VL'K1L`V'II1`'(A$,K,1)"&"(A$,K,1)"@"II1j'Kt'(40I)2~'10130'1'SPACES0'CURRENT1L'KCURRENTL'(A$,K,1)" "ĂK'WRD$(1000:J ^$(% WORDS AT BEGINNING 4&%L(A$)D0%HT(40L)2S:%X2301jD%X1X5:X1249570yN%X1S9610X%X1:" ";b%XS9590l%X:HT:A$v%6,X230:7,2:768%X% ' LOWER CAS#X1X1#Y:X#L$'#Y1:X10#" ">#X1X:Y1YE#YP#1:15_#(WRD$,T)f#Tv$T34171$2:T$"MOTION ""$J150:,$6,T:7,2:7686$@$A$"WRITTEN BY":S5:9500J$A$"LARRY SELTZER":S7:9500T$J1:779,208:780,246:781,166:782,6:783,76:784,0:785,03:786,96G^\(# INTRODUCTION b2#x<#WRD$"PROJECTILE"F#T110P#X11:Y110Z#L$(WRD$,T,1)d#TTT14n#A(TT1)22x#BTTA#Y2311#6,Y:7,2:768#XYAB (T10.5)10 X S(S10.5)10;b V((GTV0)10.5)10Ol 3:24:T:868cv 5:24:S:868w 7:24:V:868 3070 @ MUSICAL POKESJ768,173:769,48:770,192:771,136:772,208:773,4:774,198:775,7AT776,240:777,8:778,20200  15000 ) THROW UP BALL 1 T0B 54,0:55,96V 3:15:"TIME:"n 5:15:"HEIGHT: " 7:15:"VELOC.: " V030 TT.2 S12GTTV0T 0 20,191Y1& S032200 3: 20,191(S2)D Y1S2 N TING :ER(222)& DD$;"CLOSE ";NF$5 NER5ī440w XER6ĺ:"File not found. Try again. Key>";:A$::255:260 b:"DOS error #";ER;" has occurred.":Ain. Key>";:A$::255:290 :"DOS error #";ER;" has occurred.":"Y"A$"N"440 A$"N"530/ D$;"PR# 1"? S15::SJ A1N^ LL1:L26160p 10);L$(A):w A :"Please touch ESC and (Control Q) to deactivate the 80 column output =>";:A$ D$;"PR#0" &D$;"RUN MENU" 0 ERROR HANDL";NF$ 6D$;"READ ";NF$/ @L$"":X(16384)? JX127İ200G TC$V ^C$R$380c hL$L$C$l r340w |R$;D$ F1310 10)CA$;L$ :L$(N)L$:NN1 310 F1470 255::25)"Output to the printer? (Y/N> ";::A$: A$1::S L0:AA1:520- X198SPSP10A X211SPSP10R SP10SP10e SP255SP255m SP X0:16368,0:  MAIN PROGRAM :29)"Loading data. Be sure" :29)"the printer is ready." "D$;"VERIFY ";NF$ ,D$;"OPEN F11:270< x12:22)" To change the speed during this"z :22)"presentation touch ";::"F";::" for Faster or" :22);::"S";::" for Slower. Touch any key => ";:A$: 165:SP165::290 S14::S 40);P:PP1 S11 READER/PRINTER#L$(500)8CA$(1):D$(4)U(D$(4):R$(13):N1:P1f2NF$"IN.TXT"p<570FD$;"PR#3"P:28);::"**> CIRCULAR MOTION <**":Z8:28:"SCREEN OR PRINTER (S/P>";:A$:A$dA$"S"A$"P"90 nA$"P"     ALL, WE CAN USE THE FORMULA:":100001 CENTER1M A$"V = GT + V ":10000Y VTVT1u A$" O":10000 CENTER0 15000A$"@NOW LET'S SEE WHAT WILL HAPPEN WHEN WE THROW A BALL UP WITH AN INITIAL VELOCITY OF 30 METERS/SECOND":100F THE BALL, WE CAN JUST USE THE FORMULA:":10000E pVTVT1:CENTER1h zA$"S = 1/2 G T + V T":10000t VTVT3 A$" 2 ":10000 A$" O ":10000 CENTER0 VTVT1$ A$"AND TO COMPUTE THE VELOCITY OF THE B *CENTER15 4A$"@DISTANCE = @RATE X @TIME":10000B >CENTER0M H15000 RA$"@FIRST, LET'S JUST EXAMINE THE VERTICAL COMPONENT OF THE MOTION. @TO DO THIS, IMAGINE THROWING A BALL UP IN THE AIR.":10000 \VTVT10 fA$"@TO COMPUTE THE HEIGHT O VTVT1 A$"@IF THE MOTION TAKES PLACE IN A FRICTIONLESS ENVIRONMENT, THE HORIZONTAL MOTION IS FOUND USING THE VERY SIMPLE FORMULA:":10000 CENTER1 A$"S = VT":10000 CENTER0 A$"@THIS IS, OF COURSE, THE SAME FORMULA AS:":10000 (4);"BLOAD CHAR TABLE"! 8000D $VT1:SOUND0:CENTER0:WIDTH40O .G9.8U 8[ Bi  STUFF o  A$"@PROJECTILE MOTION IS SIMILAR TO FREEFALL MOTION, EXCEPT THAT IT INCLUDES A HORIZONTAL COMPONENT IN ADDITION TO A VERTICAL ONE.":10000 1000: INITIALIZE)9000:INTRO;2000: STUFFU(3000: THROW UP BALLi254,189:55,158q<:F11:12:P"PLEASE WAIT":Z(4);"RUN PRO.V" INITIALIZE 16384(768)1731060(4);"BLOAD HIRES CHAR"      ND17370!CY13"!CX40145!C6,X:7,2:768?!CX:YK!CVTVT2Q!C7,2:768 CX C17040- CSOUND173708 CY13G CX4014Z C6,X:7,2:768d CX:Yp CVTVT2v CA CG$G$(A128) lCL(G$) vCVT:L5:(A) {CSOUND17310 CX2012 C6,X:7,2:768 CX C17040!CSOUBSOUND17180BY110)BX1256<C6,X:7,3:768DC:XCVT:5:" "b&CG$""m0C17040w:CLL1DCG$(G$,L)NCVT:L6:" "XC17040bCG$G$(A128)lCL(G$)vCVT:L5:(A){CSOUND17310CX2012 C6,X: @CENTER0@'hB HIRES INPUT 2rB1:VT>|B"===>"HBG$""XBA(16384)iBA12717040wB16368,0BA14117320BA13617250BSOUND17120BX1205B6,X:7,2:768BBL(G$)BL117210BL117180:C1:L4:VTVT2.:VT:1:(CURS$,C,1)^:KEY(16384):KEY128CC1L(CL):15080l8;16368,0rL;> CLEAR PART OF SCREEN >0>X0135>X,0X,190>X>CENTER0>VT1>t@ CLEAR ENTIRE SCREEN~@@VT1E THE OBJECT IS FALLING TOWARDS THE EARTH, THE VELOCITY IS NEGATIVE.":10000Wj112670yt1A$"@THAT IS CORRECT!":10000~1150001160001: PRESS ANY KEY :16368,0:VT22:CENTER0:CURS$"!/-\":A$"! PRESS ANY KEY":10000 (G$).1(VA).812660481A0(VA).812640B1A$"@THAT IS NOT CORRECT. @THE CORRECT VALUE FOR THE FINAL VELOCITY IS:":10000L1A$(V)"M/SEC":10000V112670L`1A$"@YOU HAVE THE CORRECT MAGNITUDE, BUT YOUR DIRECTION IS INCORRECT. @SINC10000$0A$"(-9.8 M/SEC )":1000000VTVT3N0A$" 2 ":10000Z0VTVT1}0A$"INTO THE EQUATION:":100000A$"V = GT":100000150000160000A$"@WHAT IS THE FINAL VELOCITY???":100001170001160001VGTI$1A16000Op0A$"@WHAT IS THE FINAL VELOCITY (PRESS '@H' IF YOU NEED HELP)":10000Zz017000e016000}0(G$,1)"H"125700A$"@TO FIND THE FINAL VELOCITY, JUST SUBSTITUTE THE TIME":100000A$"("(TI)"SECONDS) AND":100000A$"THE GRAVITY": 0A$(TI)" SECONDS":10000)*012350J40A$"@THAT IS CORRECT":10000U>015000`H016000R0A$"@NOW THAT YOU HAVE THE TIME IT TAKES TO FALL 1000M, FIND OUT WHAT THE FINAL VELOCITY WILL BE WHEN YOUR FRIEND HITS THE GROUND.":10000\015000f0:10000/15000/16000W/A$"@NOW HOW LONG (IN SECONDS) WILL IT TAKE??":10000b/17000m/16000/T((21000G))/TI(T100)1000AN(G$) 0(TIAN).6123400A$"@SORRY, THAT'S NOT CORRECT. @THE CORRECT ANSWER IS:":10000 N/VTVT1'X/A$"S = 1/2 GT":100003b/VTVT3Ol/A$" 2":10000[v/VTVT2/A$"FOR T, AND REPLACE S AND G WITH":10000/A$"1000M AND 9.8M/SEC":10000/VTVT3/A$" 2":10000/VTVT1/A$"RESPECTIVELY"FOR YOUR FRIEND TO REACH THE GROUND.":100007/15000B/16000/A$"@HOW LONG (IN SECONDS) WILL IT TAKE? (TYPE '@H' IF YOU NEED HELP)":10000&/170000/16000:/(G$,1)"H"12270D/A$"@TO FIND THE ANSWER, JUST SOLVE THE EQUATION:":10000KE. @ONE WAY OF GETTING RID OF HIM WOULD BE BY DROPPING HIM OFF OF A 1000M HIGH TOWER.":10000j,15000u,16500{,. SECOND INTRODUCTION .VT1:WIDTH19.ANYKEY1,.A$"@NOW THAT WE HAVE THE TOWER, LET'S COMPUTE THE TIME IT WILL TAKE "AND THE EQUATION FOR DISTANCE TRAVELLED IS:":10000A,CENTER1c,A$"S = 1/2 GT + VOT":10000o,VTVT3,A$" 2 ":10000,VTVT3,15000,16500_,A$"@TO BETTER EXPLAIN THIS, IMAGINE YOU HAD A FRIEND THAT YOU DIDN'T LI000$,VTVT33.,A$" 2":10000H8,VTVT1:CENTER0B,A$"@THE VELOCITY AND DISTANCE TRAVELLED ARE NOT CONSTANT, HOWEVER.":10000L,A$"@THE EQUATION FOR VELOCITY IS:":10000V,CENTER1`,A$"V = GT":10000j,CENTER04t,A$TION OF THE OBJECT REMAINS CONSTANT, AND IS EQUAL TO THE GRAVITATION OF THE PLANET ON WHICH IT OCCURS.":10000y+15000,16500,A$"@FOR EXAMPLE, ON @EARTH, THE ACCELERATION IS ALWAYS EQUAL TO:":10000,CENTER1,A$"-9.8 METERS/SECOND ":10+A$"@FREEFALL":10000F+A$"@WRITTEN BY @LARRY @SELTZER":10000R+VTVT1_+CENTER0+A$"@FREEFALL IS A CONDITION IN WHICH THE ONLY FORCE ACTING ON AN OBJECT IS THE FORCE OF GRAVITY.":10000+VTVT1n+A$"@DURING FREEFALL, THE ACCELERA1K+KK32# +K23K23:EN1/*+K1:16?4+" "J>+K:16ZH+"FREEFALL"lM+SOUND11100R+6,K:7,10:768\+J1300:f+K1Kp+K211140z+J1300:+EN11030+J1500:J++16500+WIDTH40+CENTER1D10450(L$" "10450$(X124(S(16336);(X^(L$" "L$"."L$","10450(FINISH20150(L$" .")75(L$",")(X1FINISH:X(J(" ";(CURRENT(VTVT2(* FIRST INTRODUCTION+: +K1:K"@"NLNL1 'K/ (SPACESNLWIDTH10280; (VTVT2B (1N (SPACE0e ((SPACESSPACESNL1m 2(VTy <(J1LW F(LL32 P(L$(WRD$,J,1) Z(L$"@"L$"&"JJ1:LL064(L$"&"):10320 d(I(L$) n(I65I90LL0 x((ILL);(SOUN1)"@"II1 j'K! t'(40I)2, ~'101303 '1@ 'SPACES0Q 'CURRENT1Lb 'KCURRENTLy '(A$,K,1)" "ĂK 'WRD$(A$,CURRENT,KCURRENT) 'CURRENTK 'LW(WRD$):NLLW 'LW0SPACESSPACES1:10460 'K1(WRD$) '(WRD$,K,1)3 #X,2 #X" #X25025420 #X,5X,297 #XF #X271279T #X,5X,50[ #Xn #0,191279,191t # ' LOWER CASE PRINTER '54,0:55,96 $'L(A$) .'VT 8'CENTER010120 B'I0 L'K1L V'II1 `'(A$,K,1)"&"(A$,K,,208:780,246:781,166:782,6:783,76:784,0:785,03:786,96B ^X (# SET UP SCREEN l 2#279,191260,30 <#210,30191,191 F#260,30191,191 P#279,191210,30 Z#140,0279,0 d#140,4279,4 n#X2781405 x#X1,1:X1,1 #X1,3:X1,0 (4);"BLOAD MAN,A$7000"; (4);"BLOAD CHAR TABLE"X (4);"BLOAD HIRES CHAR"^ e 3p G9.8| SOUND0  @ MUSICAL POKES J768,173:769,48:770,192:771,136:772,208:773,4:774,198:775,7< T776,240:777,8:778,202:779w 7000: INITIALIZE,11000:INTRO 1?9000: SCREENT(12000:INTRO 2 `254,189l<55,158rFxPZ11d12nx"PLEASE WAIT"(4);"RUN FF"X INITIALIZE b16384l(768)173(786)967070 v800          .A$ .10 A'L(A$)'(40L)2);A$&$'Y1L>.'(40L)2Y1:"-";D8'J='PB'X*24^+z +"PRESS THE SPACE BAR";+ +A$*+A$" "ĺ(7):110404+.:."TO GO BACK TO THE MAIN MENU,".:"PRESS ANY KEY.".A$.10 MPUTER HAS FINISHED PLOTTING"O :"THE PROJECTILE, YOU WILL HEAR A CLICKING"| "SOUND. THIS SOUND INDICATES THAT THE" :"COMPUTER HAS FINISHED PLOTTING, AND YOU" :"SHOULD PRESS THE SPACE BAR TO RETURN" :"TO THE MENU." 12000ARAMETERS, WHICH ARE THE"Hn :"ANGLE OF INCLINATION, AND THE INITIAL"vx :"VELOCITY. THIS IS BECAUSE THEY BOTH" :"ARE EQUAL TO ZERO WHEN THE PROGRAM IS" :"RUN FOR THE FIRST TIME." 11000  A$NM$(X):10000 :"WHEN THE COITH THE"5S :"PROJECTILE. TO CHANGE A PARAMETER,"cT :"TYPE IN THE NUMBER OF THE PARAMETER,"U :"RETURN, THE CORRECTION, AND RETURN AGAIN"V 11000W X A$NM$(X):10000Z :"TO USE THE PROGRAM, YOU MUST CHANGE"d :"THE FIRST TWO P0 ( :X4.2 A$"PROJECTILE PLOTTER":10000]< :"THIS PROGRAM ALLOWS YOU TO ENTER SOME"F :"DATA ABOUT A PROJECTILE AND SEE IT"P :"PLOTTED ON THE SCREEN."Q :"WHEN YOU RUN THE PROGRAM, YOU WILL SEE"R :"A LIST OF PARAMETERS DEALING WWER AND THEN PRESS"C:"RETURN TO ALLOW THE COMPUTER TO ACCEPT"O:"IT.Z 11000`  A$"PROJECTILE MOTION TUTORIAL":10000$ :"WHEN THE TUTORIAL IS FINISHED, THE". :"PROJECTILE PLOTTER WILL BE RUN"8 :"AUTOMATICALLY"B 1200RAM. WHENEVER THE SCREEN BECOMES"S:"FULL WITH WRITING, YOU WILL BE ASKED TO":"PRESS ANY KEY. YOU CAN PRESS ANY KEY":"EXCEPT THE RESET, SHIFT, OR CTRL KEYS.":"WHEN YOU ARE ASKED TO ENTER AN ANSWER,":"PLEASE ENTER YOUR ANSTTER"3 :"WHICH WOULD YOU LIKE INFORMATION ON?"C15:"===>"R 15:5:A$^*A(A$)l4A12200z>A22600H(7)R2080:X3A$"PROJECTILE MOTION TUTORIAL":10000:"VERY LITTLE HAS TO BE DONE DURING THIS"":"PROGOGRAM WILL",:"RETURN TO THE MAIN MENU"712000=_A$"PROJECTILE MOTION":10000:"THE PROJECTILE MOTION LESSON PROGRAM IS":"BROKEN UP INTO TWO PARTS. THEY ARE::"1. PROJECTILE MOTION TUTORIAL":"2. PROJECTILE PLO DEMONSTRATION WILL BE RUN"3 J:"AUTOMATICALLY."> T12000H :X2o A$"FREEFALL DEMONSTRATION":10000 :"NOTHING AT ALL HAS TO BE DONE DURING" :"THIS DEMONSTRATION." ::"JUST SIT BACK AND ENJOY." ::"WHEN IT IS OVER, THE PRED TO ENTER AN ANSWER,"G :"PLEASE ENTER YOUR ANSWER AND PRESS THE"t :"RETURN KEY TO ALLOW THE COMPUTER TO" :"ACCEPT YOUR ANSWER." 11000 " ,A$"FREEFALL TUTORIAL":10000 6:"WHEN THE PROGRAM IS FINISHED, THE" @:"FREEFALLO BE DONE DURING THE"A :"RUN OF THIS PROGRAM. WHENEVER THE"p :"SCREEN BECOMES FULL WITH WRITING, YOU" :"WILL BE ASKED TO PRESS ANY KEY. YOU CAN" "PRESS ANY KEY EXCEPT THE RESET, SHIFT," :"OR CTRL KEYS." :"WHEN YOU ARE ASKTUTORIAL". :"2. FREEFALL DEMONSTRATION."^ $::"WHICH WOULD YOU LIKE INFORMATION ON?"o .15:"===>"; 815:5:A$:A$ BA(A$) LA11500 VA21700 `(7):1080 :X1 A$"FREEFALL TUTORIAL":10000 :"VERY LITTLE HAS T::"Touch key choice."( Z19:"===>"; _19:5:A$:A$G dA(A$)U nA11000c xA22000} A3ĺ(4);"RUN MENU" 95  A$"FREEFALL":10000 :"THERE ARE TWO PARTS TO THE" :"GRAVITY/FREEFALL LESSON:" :"1. FREEFALL  ::16368,00A$"INSTRUCTIONS":10000`(:"There are two programs in this system."2:"For which would you like information?"<:"1. Gravity/Freefall lesson"F:"2. Projectile Motion lesson"K:"3. Return to Main Menu" P        @ '3<JRXdq~ ,,$$36..$$%%..">'$$$..&%%666$$$--:?6.-= $,,-5>?<I9?'$$,$$$-557?$$$M1$";A$:TA$:T LOOK"::"IN YOUR TEXT OR SEE YOUR INSTRUCTOR.":\0v"GOOD LUCK. BYE FOR NOW!":24:10050|0:D13:(7)::20:"END"0D11500:::(4);"RUN MENU"0?0(#12:2,3737:2,3738:0':80'" ************************":1$'" :4/N"SOLVING--"::7::"Y = 3.67 METERS":::/X"THAT ALL THERE IS TO IT! TYPE 1 TO"::"REVIEW, 2 TO CONTINUE: ";A:A11570/b:6:"THAT'S THE END OF THIS PROGRAM."::Z$;", YOU SHOULD PROBABLY WORK":20l"SOME EXTRA PROBLEMS FOR PRACTICE.EMBER THAT X & Y MOTIONS DO NOT"::"AFFECT EACH OTHER, BUT THE FLIGHT TIME":.0"IS THE SAME. SO, T= 1.3 SECONDS."::10050.::3:"THEREFORE--"::" Y = V(0)Y*T - 1/2*G*T^2":/D:"SUBSTITUTING--"::" Y = 9.192*1.3 - 1/2*9.8*(1.3)^2"::"T = 1.3 SECONDS":::10050-:4:"HOW DO WE FIND THE ELEVATION WHERE THE"::"ARROW HITS THE CLIFF? ";:10050::-"HINT: HOW MUCH TIME DOES THE ARROW HAVE"::"TO GO UP AND DOWN?":-"INPUT THE ARROW'S FLIGHT TIME: ";A::N.&"REM)X = 12*COS(50) = 7.714 M/S"::" V(0)Y = 12*SIN(50) = 9.192 M/S"::10050,::"SINCE WE KNOW THE RANGE (10 METERS) AND"::"AND THE HORIZONTAL VELOCITY (7.714 M/S),","WE CAN FIND THE FLIGHT TIME--"::" T = RANGE/V(X) = 10/7.714":-5:22:" TIME MARKERS SHOW THE MOTION."H+Y501:7:203Y,11Y2:g+Y03:7:203Y,11Y2:+T11000::24:" TYPE 1 TO SEE AGAIN, 2 TO CONTINUE: ";A:A11620+::2:"HERE'S THE SOLUTION:":::" FIND V(0)X & V(0)Y--":L," V(000:1:X3035:14,36X::7:5,36j*^22:" PRESS TO SHOOT THE ARROW.";A$:T11200:::(7)y*hY50.5*r7:203Y,11Y2:0:203Y,11Y2:*|YO3.5*7:203Y,11Y2:0:203Y,11Y2:*(7):7:29,20&+T11400::ORIZONTAL AND FIRED WITH AN INITIAL":~)6"VELOCITY OF 12 M/S TOWARD A VERTICAL"::"CLIFF 10 METERS AWAY. HOW HIGH ABOVE":)@"THE GROUND DOES THE ARROW STRIKE THE"::"CLIFF?"::)J"COPY THE PROBLEM AND THEN WE'LL WATCH.":24:10050"*T::90 RANGE = V(X)*T":(" RANGE = 12 * 4"::8::"RANGE = 36 METERS"::::"TYPE 1 TO REVIEW, 2 TO CONTINUE: ";A:A11240(":3:"LET'S DO ONE MORE PROBLEM, THIS TIME"::"IN REVERSE:"::&),"AN ARROW IS AIMED 50 DEGREES ABOVE THE"::"H.4 = 0*T - 1/2*9.8*T^2"::10050:M'"THEN,"::" T^2 = 78.4/4.9":'"AND,"::5::"T = 4 SECONDS":::10050:1550':10:"GREAT WORK, ";Z$;"!"::"KEEP GOING...":22:10050(:5:"SINCE T = 4 SECONDS, WE FIND THE RANGE--"::" 1/2*G*T^2":::10050v&::"REMEMBER THAT Y = -78.4 METERS, SINCE"::"THE FINAL POSITION IS BELOW THE START."::&" INPUT THE FLIGHT TIME: ";A::A41540&:3:(7):"SORRY, ";Z$;".":::" Y = V(0)Y*T - 1/2*G*T^2":&'"OR,"::" -78S KICKED HORIZONTALLY.":?%" ENTER V(0)Y: ";A::A01460M%"SURE!";%25::"V(0)Y = 0 M/S":::"SINCE THE BALL HAS NO INITIAL VERTICAL"::"VELOCITY. ";:10050&:4:"NOW FIND THE TIME OF FLIGHT--":::"USE: Y = V(0)Y*T - HE SOLUTION. YOU SHOULD BE ABLE"s$x"TO GET THIS ONE ON YOUR OWN.":::"START BY FINDING THE X & Y COMPONENTS":$"OF THE INITIAL VELOCITY."::" ENTER V(0)X: ";A::A121430$"OF COURSE!";%25::"V(0)X = 12 M/S":::"BECAUSE THE BALL I.5,#<9:162Y,11Y2:0:162Y,11Y2:L#F(7):9:26,36:T11200:y#P22:" TIME MARKERS SHOW THE MOTION."#Z9:14,11:Y05:162Y,11Y2::T11000:#d24:" TYPE 1 TO SEE AGAIN, 2 TO CONTINUE: ";A:A11280 $n::2:"NOW FOR TMETERS/SECOND?":::"COPY THE PROBLEM. THEN WE'LL LOOK AT IT."::10050v"::9000:3:X515:12,36X::9:13,11" 22:" PRESS TO KICK THE BALL.";A$":T1800::(7)"X02:9:13X,11"(T150:T:0:13X,11:#2Y05:"HIGH.":22:10050?!:3:"HOW ABOUT ANOTHER PROBLEM?"::!"A MAN KICKS A BALL HORIZONTALLY OFF THE"::"ROOF OF A BUILDING 78.4 METERS TALL.":!"HOW FAR FROM THE BASE OF THE BUILDING"::"DOES IT LAND IF THE INITIAL VELOCITY IS":H""12 ::" Y = (173.2)^2/(2)(9.8)":U "AND,":9::"Y = 1530.5 METERS"::: "IT'S A BIG CANNON!":::"TYPE 1 TO REVIEW, 2 TO CONTINUE: ";A:A1910 1240!:10:"EXCELLENT WORK, ";Z$;"!"::"IT'S A PRETTY BIG CANNON TO SHOOT THAT":OW":T"HIGH IT GOES.":::"INPUT YOUR MAX. HEIGHT: ";A::A1530A15311230:4:(7):"SORRY. HERE'S THE CORRECT SOLUTION:":"AT MAXIMUM HEIGHT, V(Y) = 0. SO,"::" V(Y)^2 = V(0)Y^2 - 2*G*Y":' " 0^2 = (173.2)^2 - 2*9.8*Y"ROBLEM SOLUTION AGAIN, TYPE 1."::"TYPE 2 TO CONTINUE: ";A:A1910~::"AS A CONTINUATION OF THIS PROBLEM, FIND"::"THE MAXIMUM HEIGHT OF THE SHELL."::10050:3:"HINT: THE FACT THAT THE SHELL MOVES IN"::"THE X DIRECTION HAS NO EFFECT ON HEC":::VB"* NOW YOU CALCULATE THE RANGE."::" INPUT YOUR ANSWER (NO UNITS):";AmL:5:A35351120{V"GOOD!";`10::"RANGE = V(X)*T = 3535 METERS":::j"THAT'S NOT SO BAD, IS IT?"::::"IF YOU WOULD LIKE TO REVIEW THIS":Dt"P= 0 METERS ";:10050d:3:"USE THIS EQUATION:"::" Y = V(0)Y*T - 1/2*G*T^2"::10050$:"THEN,":" 0 = 173.2*T - 1/2*9.8*T^2":."AND":" -173.2*T = -1/2*9.8*T^2"::10050 8:"FINALLY,"::6::"T = 173.2/4.9 = 35.35 S;49::"V(0)Y = V(0)*SIN(60) = 173.2 M/S":::"NOW USE THE VALUE OF V(0)Y AND THE"::"EQUATIONS OF MOTION FOR GRAVITATIONAL":"ACCELERATION TO FIND THE TIME OF FLIGHT."::"REMEMBER THAT THE SHELL HITS THE GROUND,":"SO: Y = Y(0) R FOR V(0)X: ";A::A100970,"GOOD!";c10::"V(0)X = V(0)*COS(60) = 100 M/S":::10050:3:"REMEMBER THAT THIS IS THE VALUE OF THE"::"CONSTANT HORIZONTAL VELOCITY."::"INPUT YOUR VALUE FOR V(0)Y: ";A::A173.21010"YES!" WITH A MUZZLE":h"VELOCITY OF 200 M/S AT AN ANGLE OF"::"60 DEGREES ABOVE HORIZONTAL. FIND THE":"RANGE OF THE SHELL. ";:10050::"COPY THE PROBLEM. THEN TRY TO FIND THE"::"X & Y COMPONENTS OF V(0)."::"INPUT YOUR ANSWE = 0, Y(0) = 0.":gz"> THE STARTING TIME IS T(0) = 0."::"> THE ANGLE (THETA) IS MEASURED FROM ":" THE X-AXIS.":::"COPY THESE CONVENTIONS IF YOU NEED TO."::"THEN, ";:10050:3:"NOW HERE'S A PROBLEM:":::"A CANNON SHELL IS FIRED ";:10050n\:2:"THE CONVENTIONS WE USE ARE THE SAME AS"::"THOSE ESTABLISHED IN PREVIOUS PROGRAMS:"::f"> UP AND RIGHT ARE POSITIVE."::"> DOWN AND LEFT ARE NEGATIVE.":p"> THE STARTING POSITION (EVEN IF ABOVE"::" THE GROUND IS X(0)0)X!"::" V(0)Y = V(0)*SIN(THETA)"z>" NOTE: V(Y) WILL CHANGE!"::" 4. APPLY EQUATIONS OF MOTION IN THE"H" Y DIRECTION TO FIND THE TIME OF":" FLIGHT (T)": R" 5. FIND RANGE = V(X) * T"::"COPY THE METHOD FOR REFERENCE:";:10050?:2:"METHOD:"::" 1. ESTABLISH CONVENTIONS": " 2. DETERMINE KNOWN & UNKNOWN VALUES"::" 3. FIND THE X & Y COMPONENTS OF THE"*" INITIAL VELOCITY--"::" V(0)X = V(0)*COS(THETA)")4" NOTE: V(X) = CONSTANT V("THE PROJECTILE. YOU ARE TO SOLVE FOR THE":"RANGE (MAX. HORIZONTAL DISTANCE)."::"OF COURSE, THE PROBLEM CAN BE CHANGED"::"AROUND (GIVEN RANGE AND ANGLE, FIND THE":  "INITIAL VELOCITY). BUT, LET'S START WITH":"THE EASY ONE! Z$;"?"::"TYPE 1 TO SEE AGAIN, 2 TO CONTINUE: ";A:A1370:3:"NOW LET'S LOOK AT A METHOD FOR SOLVING"::"PROJECTILE PROBLEMS. ";:10050::"IN THE BASIC PROBLEM, YOU ARE GIVEN THE"::"INITIAL VELOCITY AND ANGLE (THETA) OF":SND BY THE"::"ACCELERATED MOTION IN THE Y DIRECTION."::"THEREFORE, THE TIME IT TAKES THE PRO-"::"JECTILE TO GO UP AND DOWN IS EXACTLY":"THE AMOUNT OF TIME THE PROJECTILE HAS"::"TO MOVE SIDEWAYS."::;"DO YOU WANT TO WATCH AGAIN, ";00:T@11:10,36203Y:7:21,3611Y2:1:203Y,11Y2:}24:" TYPE 1 TO SEE AGAIN, 2 TO CONTINUE: ";A:A1620::3:"YOU SHOULD SEE THAT THE PROJECTILE"::"POSITION IS DETERMINED BY THE CONSTANT"::"VELOCITY IN THE X DIRECTION Ab24:" TYPE 1 TO SEE AGAIN, 2 TO CONTINUE: ";A:A1370l::9000:1:5,36:22:" HERE IT IS IN SLOW MOTION.":T11500:vY501:T11000:T11:10,36203Y:7:4,2011Y2:1:203Y,11Y2:7:21,3611Y15:T110:Y50.5<13:203Y,11Y2:0:203Y,11Y2:J&YO5.5u013:203Y,11Y2:0:203Y,11Y2::13:35,36DT11000::22:" TIME MARKERS SHOW THE MOTION."NY501:13:203Y,11Y2:XY05:13:203Y,11Y2:;5!7:4,11Y2:0:4,11Y2:/Y05.5X7:4,11Y2:0:4,11Y2::7:4,36T11000::22:" TIME MARKERS SHOW THE ACCELERATION."7:Y05:4,11Y2::24:10050:22:" NOW COMBINE THE MOTIONS."13:5,36:T1100000: Y010.1? 11:53Y,10:0:53Y,10::11:35,10 T11000::22:" TIME MARKERS SHOW CONSTANT VELOCITY. " 11:Y010:53Y,10::24:10050 :22:" HERE'S A VERTICAL ACCELERATION." 7:4,36:T11000:Y50. MOTION AS TWO":X ^"SEPERATE MOTIONS WHICH LAST FOR THE"::"SAME AMOUNT OF TIME.": h:"LET'S SEE HOW THE MOTIONS COMBINE."::"WATCH CAREFULLY, ";Z$;"!":::10050 r::9000:22:" HERE'S A CONSTANT HORIZONTAL VELOCITY." |11:5,10:T110UE TO"( ," GRAVITY"::3 610050 @:3:"THESE TWO MOTIONS OCCUR AT THE SAME"::"TIME, AND THE RESULT IS PARABOLIC": J"PROJECTILE MOTION. ";:10050:: T"THE TRICK IN SOLVING PROBLEMS IS TO"::"CONSIDER THE PROJECTILEX "PROJECTILE MOTION CAN BE CONSIDERED AS"::"TWO SIMULTANEOUS MOTIONS WHICH ARE":o :"INDEPENDENT": 12:13:"EXCEPT FOR THE TIME FACTOR.":::"THE MOTIONS ARE:": "" 1. HORIZONTAL - CONSTANT VELOCITY"::" 2. VERTICAL - ACCELERATION DFFICULTY WITH":_ "THEM--SO PAY CLOSE ATTENTION.":::"YOU SHOULD HAVE ALREADY WATCHED:": :"MOTION AND GRAVITY":::"WHEN YOU'RE READY, ";:10050 :3:"THE BASIC IDEA IS SIMPLE, BUT YOU MUST"::"UNDERSTAND IT WELL TO SOLVE PROBLEMS."::m)d- PROJECTILE MOTION%n10000[:3:"HELLO!"::"PLEASE TELL ME YOUR NAME: ";Z$8:"THANK YOU, ";Z$;". THIS PROGRAM"::"IS DEVOTED TO THE SOLUTION OF PROJECTILE" "PROBLEMS. ALTHOUGH THEY'RE REALLY NOT "::"TOO HARD, STUDENTS HAVE DI""" " " " " """""""""!!! ! ! ! ! !!!!!!!!!