a
(a) Time taken by first stone to reach the water surface from the bridge be \(t\), then

\(h = ut + \frac{1}{2}g{t^2} \Rightarrow 44.1 = 0 \times t + \frac{1}{2} \times 9.8{t^2}\)

\(t = \sqrt {\frac{{2 \times 44.1}}{{9.8}}} = 3\;sec\)

Second stone is thrown 1 sec later and both strikes simultaneously. This means that the time left for second stone

\( = 3 - 1 = 2\;sec\)

Hence \(44.1 = u \times 2 + \frac{1}{2}9.8{(2)^2}\)

\( \Rightarrow 44.1 - 19.6 = 2u \Rightarrow u = 12.25\;m/s\)