Questions
A dynamic blast blows a heavy rock straight up with a launch velocity of 160 m/sec. It reaches a height of after t sec. The velocity
of the rock when it is 256 m above the ground on the way up is
detailed solution
Correct option is B
v=dsdt=160−32t. We now find values of t for which s(t)=256. So 160t−16t2=256⇒16t2−10t+16=0⇒(t−2)(t−8)=0)⇒ t=2,t=8 So v(2)=160−32.2=96 v(8)=160−256=−96 So __ the velocity on the way up in 96m/s.Talk to our academic expert!
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