A bulletfired into a fixed target loses half of its velocity after penetrating 3cm. How much further it will penetrate before coming to rest assuming that it faces constant resistance to motion?  

 let the velocity of th bullet Vand after penetrating 3cm it got the velocity V/2.

now,(V/2)^2=V^2-2fs

    or,f=3V^2 /8s=V^2 /8

a another equeation when V=0,then 0=(V/2)^2-2fs'

or,s'=(V/2)^2 *1/2f

or,s'=1(where f=V^2/8)

Ans:it will penetrate further 1cm.