A bullet fired horizontally into a thick wooden block undergoes...

Let the entry velocity of the bullet be $u$ and the final velocity after travelling $30\text{ cm}$ be $v$.

Given:
Distance $s = 30\text{ cm} = 0.3\text{ m}$
Time $t = 0.1\text{ s}$
Final velocity $v = \dfrac{u}{2}$

For uniformly accelerated (or decelerated) motion, the displacement can be expressed in terms of average velocity:
$s = \left(\dfrac{u + v}{2}\right)t$

Substituting the given values:
$0.3 = \left(\dfrac{u + \dfrac{u}{2}}{2}\right) \times 0.1$

$0.3 = \left(\dfrac{3u}{4}\right) \times 0.1$

$3 = \dfrac{3u}{4}$

$u = 4\text{ m s}^{-1}$

The entry velocity of the bullet is $4\text{ m s}^{-1}$.

Answer: $4\text{ m s}^{-1}$