A bullet fired into a fixed target loses half | vedclass

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(B) Let the initial velocity be $V$ and the constant retardation be $a$. Using the equation of motion $v^2 = u^2 + 2as$,where $v = V/2$,$u = V$,and $s

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$1$

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(B) Let the initial velocity be $V$ and the constant retardation be $a$. Using the equation of motion $v^2 = u^2 + 2as$,where $v = V/2$,$u = V$,and $s = 3 \ cm$: $(V/2)^2 = V^2 - 2a(3) \implies V^2/4 = V^2 - 6a \implies 6a = 3V^2/4 \implies a = V^2/8$. Now,let the total distance covered be $S$. When the bullet comes to rest,final velocity $v = 0$: $0^2 = V^2 - 2aS \implies V^2 = 2(V^2/8)S \implies V^2 = (V^2/4)S \implies S = 4 \ cm$. The additional distance covered is $S - 3 \ cm = 4 \ cm - 3 \ cm = 1 \ cm$.

$A$ bullet fired into a fixed target loses half of its velocity after penetrating $3 \ cm$. How much further will it penetrate before coming to rest,assuming that it faces constant resistance to motion? (in $cm$)

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