A bullet is fired into a fixed target looses one third of its...

When the bullet just enters the target, its equation of motion is given by

${v}^{2} - {u}^{2} = 2aS . . . \left(1\right)$

Equation (1) implies that

${\left(\dfrac{2u}{3}\right)}^{2} = {u}^{2} + 2\left(- a\right)\left(4 \times {10}^{- 2}\right) ⇒ \dfrac{4{u}^{2}}{9} = {u}^{2} - 2a\left(4 \times {10}^{- 2}\right) ⇒ \dfrac{5{u}^{2}}{9} = 2a\left(4 \times {10}^{- 2}\right) \ldots \left(2\right)$

When the bullet stops inside the target, its equation of motion can be written as

$0 = {\left(\dfrac{2u}{3}\right)}^{2} + 2\left(- a\right)\left(x\right) . . . \left(3\right)$

Equation (3) implies that

$\dfrac{4{u}^{2}}{9} = 2ax . . . \left(4\right)$

Dividing equation (2) by equation (4), we have

$\dfrac{5}{4} = \dfrac{4 \times {10}^{- 2}}{x} ⇒ x = \dfrac{16}{5} \times {10}^{- 2} ⇒ x = 3 . 2 \times {10}^{- 2}m = 32 \times {10}^{- 3}m$

Hence, $D = 32$.