b
displacement of the first stone when second is dropped \(=5 m\),

time difference in dropping of the stones is \(t =\frac{2 h }{ g }=1\; s\)

\(h =\frac{1}{2} g t ^{2}=\frac{1}{2} g ( t -1)^{2}+25 \Rightarrow t =3 \;s\)

thus, \(h =\frac{1}{2} g t ^{2}=\frac{1}{2} \times 10 \times (3)^{2}= 4 5 \;m\)