ProjectileMotion-HorizontallaunchesThescaleontheleftis1cm=5mDrawthepositionsofthedroppedballeverysecondthatitisintheair.Neglectairresistanceandassumeg=10m/s2.EstimatethenumberofsecondsthattheballisintheairFourpositionsofaballthrownhorizontallywithoutgravityareshown.Drawthepositionsoftheballwithgravity.Describethepathtakenbytheball.Howismotionintheverticaldirectionaffectedbymotioninthehorizontaldirection?TheyareindependentofeachotherItisparabolicProjectileMotionRules:1.VerticalMotioniseffectedbygravity2.HorizontalmotionisconstantMotioninthehorizontaldirectionisindependentofthemotionintheverticaldirection.4.AssumenoairresistancePhysicalExpressionsHorizontalVerticalvHi

=

vHf

=vHdH

=

vH

ta=0m/s2vvf

=vVi

+gtdv

=vvi

t+1/2gt2a=g=-9.8m/s2vHi

=vi

cosvvi

=visinExampleProblemAprojectileislaunchedfromaheightof44.1mwithainitialhorizontalspeedof20m/s.a)Howlongisittheair?b)howfardoesittravelhorizontallybeforeithitstheground?vvi

=0m/svH

=20m/s

dv

=-44.1mg=-9.8m/s2

a)dv=vvi

t+1/2gt2

-44.1m

=(0m/s)t+1/2(-9.8m/s2)t2t=

(2(-44.1m)/-9.8m/s2)=3sb)dH=vH

tdH

=(20m/s)(3s)=60m20m/s-44.1mExampleContinuedc)Whatvelocitydoestheprojectilehittheground?vvi

=0m/svH

=20m/s

dv

=-44.1mg=-9.8m/s2

t=3svVf

=vVi

+gtvVf

=(0m/s)+(-9.8m/s2)(3s)=-29.4m/sv=(vvf2+vH2)=(-29.4m/s)2+(20m/s)2=35.6m/s=tan-1(vvf

/vH)belowthehorizontal=tan-1(29.4m/s/20m/s)v=36m/s@560belowthehorizontal=55.80belowthehorizontalvHvvfvExampleContinuedvHvvfvBottomor…cansolveforspeedusingconservationofenergyKETop

+GPETop

=KEBottom

+GPEBottom1/2mvTop2+(-mghTop)=1/2mvBottom2+(-mghBottom)vTop=20m/s

hTop=44.1m

hBottom=0m

g=-9.8m/s2

vBottom=?

1/2vTop2+(-ghTop)=1/2vBottom2+(-ghBottom)1/2(20m/s)2+(-(-9.8m/s2)(44.1m)=