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.