A bullet fired into a trunk of a tree loses $1/4$ of its kinetic energy in travelling a distance of 5cm. Before stopping it travels a further distance of

Assuming the deceleration of the bullet to be constant, 

${\mathrm{v}}^2-{\mathrm{u}}^2=2\mathrm{as}$

Given the KE of the bullet dropped by 1/4th of the initial KE through a distance of 5cm

$\Rightarrow \frac{3}{4}{\mathrm{u}}^2-{\mathrm{u}}^2=2\mathrm{a}\left(5\right)$

$\Rightarrow \mathrm{a}=-\frac{{\mathrm{u}}^2}{40}$

Distance travelled before KE becomes zero is

$0-\frac{3}{4}{\mathrm{u}}^2=2\left(\frac{-{\mathrm{u}}^2}{40}\right)\mathrm{s}$

$\Rightarrow \mathrm{s}=15\mathrm{cm}$