Copyright © 2011 Pearson, Inc. 6.3 Parametric Equations and Motion.
Section 6.3 Parametric Equations and Motion
2
Section 6.3 Parametric Equations and Motion o Read page 522 o Go through examples 2, 3, and 4 starting on page 523 Projectiles This is the main topic for the section. Projectiles are things (that have no way to propel themselves) that are launched into the air like rocks, balls, arrows, etc. A golf ball hit into the air would travel in a straight line forever if not for gravity and air friction. We will ignore friction (which is not very realistic, but makes the problems easier to solve) so the only force affecting our projectiles will be the Earth’s gravity. The pull of gravity causes projectiles to travel in a parabolic arc that can be modeled with the following parametric equations: = ! cos = − 1 2 ! + ! sin + ! Where = distance traveled by the projectile in the horizontal direction = the height of the projectile = the Earth’s gravity constant (32 feet per second per second) ! = the original velocity of the projectile = the launch angle (angle of elevation) ! = the original height of the projectile = time since launch
Transcript of Section 6.3 Parametric Equations and Motion
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.
