$h = ut + \frac{1}{2}g{t^2} \Rightarrow 44.1 = 0 \times t + \frac{1}{2} \times 9.8{t^2}$
$t = \sqrt {\frac{{2 \times 44.1}}{{9.8}}} = 3\;sec$
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$
$\begin{array}{|c|c|c|c|c|} \hline t( s ) & 0 & 1 & 2 & 3 \\ \hline x ( m ) & -2 & 0 & 6 & 16 \\ \hline \end{array} $