When the bullet just enters the target, its equation of motion is given by
v2−u2=2aS...(1)
Equation (1) implies that
(32u)2=u2+2(−a)(4×10−2)⇒94u2=u2−2a(4×10−2)⇒95u2=2a(4×10−2)…(2)
When the bullet stops inside the target, its equation of motion can be written as
0=(32u)2+2(−a)(x)...(3)
Equation (3) implies that
94u2=2ax...(4)
Dividing equation (2) by equation (4), we have
45=x4×10−2⇒x=516×10−2⇒x=3.2×10−2m=32×10−3m
Hence, D=32.