Exercises 3.4Using equation (3.11) and noting thatdv/dt=(dv/dx)(dx/dt)=(dv/dx)v, we can determinethe maximum height attained by the shell.With the above substitution, equation (3.11)becomesmvdvdx=−(mg+0.1v2),v(0) = 500.Using separation of variables and integration, we getvdv10mg+v2=−dx10m⇒12ln(10mg+v2)=−x10m+C⇒10mg+v2=Ke−x/(5m).Settingv= 500 whenx= 0, we FndK=e0(10(3)(9.81) + (500)2)= 250294.3.Thus the equation of velocity as a function of distance isv2+10mg= (250294.3)e−x/(5m).The maximum height will occur when the shell’s velocity is zero, thereforexmaxisxmax=−5(3) ln±0 + 10(3)(9.81)250294.3²≈101.19 (meters).15.The total torque exerted on the ﬂywheel is the sum of the torque exerted by the motor and
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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.