A bullet fired into a wooden block loses 75% of its velocity after penetrating 15 cm into it' How much further distance it has to penetrate to be stopped completely ?

Let u be the initial velocity of the bullet and a be the retardation of the wooden board assumed to be constant. Using the formula v² =u² +2as, we have
$\begin{array}{l}{\left(\frac{\mathrm{u}}{4}\right)}^2={\mathrm{u}}^2-2\mathrm{a}\times 15\\ \mathrm{or}\frac{{\mathrm{u}}^2-16{\mathrm{u}}^2}{16}=-30\mathrm{a}\end{array}$
$\therefore {\mathrm{u}}^2=32\mathrm{a}...\left(1\right)$
Let it travel a further distance s before it is stopped. Then
$0={\left(\frac{\mathrm{u}}{4}\right)}^2-2\mathrm{as}$
or u² = 32 as … (2)
From eqs. (1) and (2).
32 a s = 32 a or s = 1 cm