section 6.3 parametric equations and motion
TRANSCRIPT
Section6.3ParametricEquationsandMotion
o Readpage522
o Gothroughexamples2,3,and4startingonpage523Projectiles
Thisisthemaintopicforthesection.Projectilesarethings(thathavenowaytopropelthemselves)thatarelaunchedintotheairlikerocks,balls,arrows,etc.Agolfballhitintotheairwouldtravelinastraightlineforeverifnotforgravityandairfriction.Wewillignorefriction(whichisnotveryrealistic,butmakestheproblemseasiertosolve)sotheonlyforceaffectingourprojectileswillbetheEarth’sgravity.Thepullofgravitycausesprojectilestotravelinaparabolicarcthatcanbemodeledwiththefollowingparametricequations:
𝑥 = 𝑉! cos𝜃 𝑡
𝑦 = −12𝑔𝑡
! + 𝑉! sin𝜃 𝑡 + 𝑦!
Where𝑥 =distancetraveledbytheprojectileinthehorizontaldirection𝑦 =theheightoftheprojectile𝑔 =theEarth’sgravityconstant(32feetpersecondpersecond)𝑉! =theoriginalvelocityoftheprojectile𝜃 =thelaunchangle(angleofelevation)𝑦! =theoriginalheightoftheprojectile𝑡 =timesincelaunch
ExampleProblem
1. Supposeabaseballishitfromaheightof3feetatanangleof32°withthehorizontal.Theinitialvelocityoftheballis120feetpersecond.
First,let’ssetuptheparametricequationsthatmodelthissituation:
𝑥 = 120 cos 32° 𝑡
𝑦 = −16𝑡! + 120 sin 32° 𝑡 + 3
a) Findtheheightoftheballattime𝑡 = 1.8
𝑦 = −16 1.8 ! + 120 sin 32° 1.8 + 3 = 65.62256 𝑓𝑒𝑒𝑡
b) Howlongwilltheballbeintheair?
Notethatwhentheballlands𝑦 = 0.
0 = −16𝑡! + 120 sin 32° 𝑡 + 3
Thus𝑦 = −0.0466and𝑦 = 4.0210(viaPlySmlt2)
Weonlycareaboutthepositiveanswer
c) Howfarawayfromthebatterwilltheballland?
𝑥 = 120 cos 32° 4.0210 = 409.20 𝑓𝑒𝑒𝑡
d) Whatisthemaximumheightattainedbytheball?
Notethatthemaximumheightofaparabolaisatthevertexwhere𝑡 = − !!!
Sointhiscase𝑡 = − !"# !"# !"°
! !!"= 1.987197
So𝑦!"# = −16 1.987197 ! + 120 sin 32° 1.987197 + 3 = 66.183 𝑓𝑒𝑒𝑡
e) Willtheballcleara10-foothighfence380feetfromthebatter?
380 = 120 cos 32° 𝑡
𝑡 = 3.73406 𝑠𝑒𝑐𝑜𝑛𝑑𝑠
𝑦 = −16 3.73406 ! + 120 sin 32° 3.73406 + 3 = 17.358 𝑓𝑒𝑒𝑡
Sotheballwillclearthefence.