A bullet from a gun is fired on a rectangular wooden block with...

For the first part of the motion of the bullet into the block, it can be written that

$\begin{cases}{\left(\dfrac{u}{3}\right)}^{2} & = & {u}^{2} - 2a \times 24 \\ & ⇒ & 48a = \dfrac{8{u}^{2}}{9} . . . \left(1\right)\end{cases}$

For the second part of the motion of the bullet through the block, it can be written that

$\begin{cases}0 & = & {\left(\dfrac{u}{3}\right)}^{2} - 2aS \\ & ⇒ & 2aS = \dfrac{{u}^{2}}{9} . . . \left(2\right)\end{cases}$

Divide equation (2) by equation (1) to obtain the distance travelled by the bullet during the second part of its motion.

$\begin{cases}\dfrac{2aS}{48a} & = & \dfrac{\dfrac{{u}^{2}}{9}}{\dfrac{8{u}^{2}}{9}} \\ & ⇒ & S = 3\end{cases}$

Hence, the total length of the block is given by

$\begin{cases}L & = & \left(3 + 24\right)cm \\ & = & 27cm\end{cases}$