A bullet when fired into a target loses half of its velocity after penetrating 20 cm. Further distance of penetration before it comes to rest is
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A useful shortcut for this specific problem: If a bullet loses \(1/n\) of its velocity after distance \(s\), it travels a further distance of \(s / (n^2 - 1)\) before stopping. Here, \(n=2\), so distance = \(20 / (2^2 - 1) = 20/3\).