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Q.

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

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a

98 m/s

b

96 m/s

c

104 m/s

d

48 m/s

answer is B.

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Detailed Solution

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.
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