If a body looses half of its velocity on penetrating 3 cm in a...

Let the initial velocity of the body be v. Hence the final velocity = v/2 Applying $v^2=u^2-2 a s \Rightarrow\left(\frac{v}{2}\right)^2=v^2-2 \cdot a \cdot 3 \Rightarrow a=v^2 / 8$ In IInd case when the body comes to rest, final velocity = 0, initial velocity = v/2 Again, $(0)^2=\left(\frac{v}{2}\right)^2-2 \cdot \frac{v^2}{8} \cdot s ;$ or $s=1 \mathrm{~cm}$ So the extra penetration will be $1 \mathrm{~cm}$