Projectile in an Inclined Plane 2.docx
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ProjectileinanInclinedPlane2
PROJECTILEINANINCLINEDPLANESOLUTIONS.
SOLUTION#1:
PhysicsforScientistsandEngineers(3rd)
ByDouglasC.Giancoli
Chapter3- Problem49
Wechooseacoordinatesystemwiththeoriginatthebaseofthehill,withxhorizontalandyvertical
withthepositivedirectionup.
Whentheobjectlandsonthehill,y=xtanφ.Thedistanceupthehillisgivenby
d2=x2+y2=x2(1+tan2φ).
Thustomaximized,wecanmaximizex.Forthehorizontalandverticalmotionswehave
x=x0+v0xt=0+(v0cosθ)t=(v0cosθ)t;
y=y0+v0yt+½ayt2=0+(v0sinθ)t+½(–g)t2=(v0sinθ)t–½gt2.
Wecombinetheseequationstogetxafunctionofθ:
y=xtanφ=(v0sinθ)(x/v0cosθ)–½g(x/v0cosθ)2,whichgives
x=(2v02cosθ/g)(sinθ–cosθtanφ).
x=(2v02/g)(cossinθ–cos2θtanφ).
Wecansimplifythisbyusingtrigonometricidentitiestogetfunctionsof2θ:
sincosθ=½sin2θandcos2θ=½(1+cos2θ)
x=(v02/g)(sin2θ+cos2θtanφ+tan)
x=(v02/g)(1/cos)[sin2θ–(1+cos2θ)tanφ].
Thelaststephasanticipated;theterminsquarebracketsontherightisseentobe(sin2+),whichallowsustoexpresstherangeas:
x=(v02/g)(1/cos)[sin(2θ+)+sinφ].
Tofindthevalueofθformaximumx,wesetdx/dθ=0:
dx/dθ=(v02/g)[2cos2θ–(–2sin2θ)tanφ]=0,whichgives
=(v02/g)[2cos2θ+2sin2θ)tanφ]=0
2sin2θ/2cos2θtanφ=0
tan2θ=–cotφ,orθ=½tan–1(–cotφ).
Notethatthenegativesignmeansusingtheanglegreaterthan90°fortheinversetangent.
SOLUTION#2:
UniversityPhysicswithModernPhysics(11th)
ByHughD.Young,RogerA.Freedman
Chapter3 - Problem89
a)Accelerationduetogravityisg=9.8m/s2.Wecanstartbywritingtheequationsforxandy
asafunctionoftime.Bynow,wecaneasilyseethatwehave:
x=x0+v0xt=0+(v0cosφ)t=(v0cosφ)t;
y=y0+v0yt+½ayt2=0+(v0sinφ)t+½(–g)t2=(v0sinφ)t–½gt2.
Wecancombinetheseequationstogetarelationbetweenxandyforpointsonthetrajectory;
Fromthefirst,wehave
t=x/(v0cosφ);
andputtingthisintothesecondonegives:
y=(v0sinφ)(x/v0cosφ)–½g(x/v0cosφ)2
y=xtanφ–gx2/2(v02cos2φ)
Whatistheconditionforthetimethattheprojectilehitstheslope?
Unlikeproblemswhereaprojectile
impactswiththeflatgroundorthewall,wedonotknowthevalueofxandyatimpact.Butsincetheincline
hasaslopeoftan+.Thesetworelationsbetweenxandyallowsustosolveforthevaluesofxandywhere
theimpactoccurs.Substituting,foryabove,wefind:
xtan=xtanθ+φ–gx2/2(v02cos2θ+φ)
Alittlearrangementgives:
gx2/2v02cos2θ+(tanθ-tanθ+)=0
Andthesolutionforxis:
x=2v02cos2θ+(tanθ+φ–tanθ)/g
Theproblemhasustosolveforthedistanceduptheslope;thisdistanceisrelatedtotheimpactvalueofx
by:
d=x/cos
d=x/cos=2v02cos2θ+(tanθ+φ–tanθ)/gcos
b)Formaximumrange,wehavedx/d=0
d/d=2v02cos2θ+tanθ+φ–tanθcos2+/gcos=0
Since
cos2θ+tanθ+φ=cos2θ+sinθ+φ/cosθ+φ
=½(2sinθ+cosθ+φ)
=½sin2(θ+φ)
d/d=2v02/gcos½sin(2θ+2φ)–tanθcos2+=0
=2v02/gcos½sin(2θ+2φ)2+tanθ2cosθ+φsinθ+φ=0
=2v02/gcoscos(2θ+2φ)+tanθsin(2θ+2φ)=0
=cos(2θ+2φ)+tanθsin(2θ+2φ)=0
–tan=cos(2θ+2φ)/sin(2θ+2φ)
cot(/2+)=cot(2θ+2φ)
/2+=2θ+2φ
2φ=/2-
φ=/4-/2
SOLUTION#3:
PhysicsforScientistsandEngineers(6th)
ByRaymondA.Serway,JohnW.Jewitt
Chapter4 - Problem50
a)Accelerationduetogravityisg=9.8m/s2.Wecanstartbywritingtheequationsforxandy
asafunctionoftime.Bynow,wecaneasilyseethatwehave:
x=x0+v0xt=0+(v0cosφ)t=(v0cosφ)t;
y=y0+v0yt+½ayt2=0+(v0sinφ)t+½(–g)t2=(v0sinφ)t–½gt2.
Wecancombinetheseequationsweget
y=xtan–gx2/2(v02cos2)
Settingx=dcos(),andy=sin(),wehave;
dsin()=tan()(dcos()–g(dcos()2/2(v02cos2).
Solvingfordgives;
d=2v02cos2sin()cos()–sin()cos()/g2cos2().Or
d=2v02cos2sin(–)/g2cos2().
b)Weeliminatet=
SOLUTION#4:
a)Wehave:
x=x0+v0xt=0+(v0cos)t=(v0cos)t;
y=y0+v0yt+½ayt2=0+(v0sin)t+½(–g)t2=(v0sin)t–½gt2.
Wecancombinetheseequationstogetarelationbetweenxandyforpointsonthetrajectory;
Fromthefirst,wehave
t=x/(v0cos);
andputtingthisintothesecondonegives:
y=(v0sin)(x/v0cos)–½g(x/v0cos)2
y=xtan–gx2/2(v02cos2)
Calltheoriginofcoordinatesthelaunchpointwithup+yandhorizontallyacrosstheincline+xThemotionoftheprojectilewith0=+φ.Theprojectilestrikestheplanewithy=xtan.Then
xtan=xtan(+φ)-gx2/2vo2cos2(+φ).
Theequationhasthetriviailsolutionx=0,andthesolutionofinterest
x=2v02/gcos2(θ+φ)(tan(+φ)–tan()).
Nowthedistancesuptheinclinecorrespondingtothexis:
x/s=cos()s=x/cos().
So,aftersimplifyingabit,
s=2v02/gcos2(θ+φ)/cos()(sin(θ+φ)cos(θ)-sin(θ)cos(θ+φ)/cos(+φ)cos()
s=2v02/gcos(θ+φ)/cos2()(sin(θ+φ)cos()–sin()cos(θ+φ)/cos2()
Atrigonometricidentity;givesus:
s=2v02/gcos(θ+φ)sin(φ)/cos2()
b)Theoptimumφisdeterminedbysettingthederivativeofswithrespecttoφtozero:
ds/dφ=2v02/g–sin(θ+φ)sin(φ)+cos(θ+φ)cos(φ)/cos2()=0
=–sin(θ+φ)sin(φ)+cos(θ+φ)cos(φ)=0
Atrigonometricidentityreducesthisto:
cos(+2φ)=0+2φ=90ºφ=45º–/2
SOLUTION#5:
a)Wehave:
x=Rcos()=v0cos()tandy=Rsin()=v0sin()t–½gt2.
Wecancombinetheseequationstogetarelationbetweenxandyforpointsonthetrajectory;
Fromthefirst,wehave
t=R/(v0cos);
Solvetheequationfortandsubstitutethisintothesecondequation
Rsin()=v0sin()Rcos()/(v0cos)–½g()Rcos()/(v0cos)2.
sin()=sin()cos()/cos()–gRcos2()/2v02cos2()
tan()=tan()–gRcos2()/2v02cos2()
gRcos()=2v02cos2()tan()–tan()
R=2v02cos2()tan()–tan()/gcos()
http:
//www.atmosp.physics.utoronto.ca/people/codoban/PHY138/Mechanics/ch3p43.pdf
http:
//academic.udayton.edu/LenoPedrotti/examples/206kinematics.pdf