Chapter 3The negative sign is present because air resistance acts in opposition to the motion of theobject. Therefore the net force acting on the shell isF=F1+F2=−mg−(0.1)v2.We now apply Newton’s second law to obtainmdvdt=−±mg+(0.1)v2².Because the initial velocity of the shell is 500 m/sec, a model for the velocity of the risingshell is expressed as the initial-value problemmdvdt=−±mg.1)v2²,v(t= 0) = 500,(3.11)whereg=9.81. Separating variables, we getdv10mg+v2=−dt10mand soZdv10mg+v2=−Zdt10m⇒1√10mgtan−1³v√10mg´=−t10m+C.Settingm=3,g.81 andv= 500 whent= 0, we FndC=1
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This note was uploaded on 03/29/2010 for the course MATH 54257 taught by Professor Hald,oh during the Spring '10 term at University of California, Berkeley.