\[(a)\;4\;cm\quad (b)\;3\; cm \quad (c)\;2\; cm \quad (d)\;1\;cm\]

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Let 'u' be initial velocity

$\bigg(\large\frac{u}{2}\bigg)^2$$-u^2=-2a \times 3$

Since it loses half its velocity after 3 cm

or $a=\large\frac{u^2}{8}$

Let x be the further distance to which it penetrates

Therefore Total distance $=x'=3+x$

$0-u^2=-2 \large\frac{u^2}{8}$$x'$

$x'=4$

Therefore $x=4-3=1\; cm$

Hence d is the correct answer.

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