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1. A dynamite blast blows a heavy rock straight up with a launch velocity of 140 ft sec (about 109mph). It reaches a height of |
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Answer» Answer AnswerWe KNOW, v= AnswerWe know, v= dt AnswerWe know, v= DTDS AnswerWe know, v= dtds AnswerWe know, v= dtds =160−32t. AnswerWe know, v= dtds =160−32t.We now FIND values of t for which s(t)=256 So AnswerWe know, v= dtds =160−32t.We now find values of t for which s(t)=256 So160t−16t AnswerWe know, v= dtds =160−32t.We now find values of t for which s(t)=256 So160t−16t 2 AnswerWe know, v= dtds =160−32t.We now find values of t for which s(t)=256 So160t−16t 2 =256 AnswerWe know, v= dtds =160−32t.We now find values of t for which s(t)=256 So160t−16t 2 =256⇒16(t AnswerWe know, v= dtds =160−32t.We now find values of t for which s(t)=256 So160t−16t 2 =256⇒16(t 2 AnswerWe know, v= dtds =160−32t.We now find values of t for which s(t)=256 So160t−16t 2 =256⇒16(t 2 −10t+16)=0 AnswerWe know, v= dtds =160−32t.We now find values of t for which s(t)=256 So160t−16t 2 =256⇒16(t 2 −10t+16)=0⇒(t−2)(t−8)=0 AnswerWe know, v= dtds =160−32t.We now find values of t for which s(t)=256 So160t−16t 2 =256⇒16(t 2 −10t+16)=0⇒(t−2)(t−8)=0⇒t=2, t=8. AnswerWe know, v= dtds =160−32t.We now find values of t for which s(t)=256 So160t−16t 2 =256⇒16(t 2 −10t+16)=0⇒(t−2)(t−8)=0⇒t=2, t=8.So v(2)=160−32.2=96 AnswerWe know, v= dtds =160−32t.We now find values of t for which s(t)=256 So160t−16t 2 =256⇒16(t 2 −10t+16)=0⇒(t−2)(t−8)=0⇒t=2, t=8.So v(2)=160−32.2=96v(8)=160−256=−96 AnswerWe know, v= dtds =160−32t.We now find values of t for which s(t)=256 So160t−16t 2 =256⇒16(t 2 −10t+16)=0⇒(t−2)(t−8)=0⇒t=2, t=8.So v(2)=160−32.2=96v(8)=160−256=−96So the velocity on the way up in 96 m/s. |
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