A stone is dropped into the water from a bridge 44.1 m above the water. Another stone is thrown vertically downward one second later. Both strike the water simultaneously, then the initial speed of the second stone is: 

$\mathrm{s} = \mathrm{ut} + \frac{1}{2}{\mathrm{at}}^2$

Acceleration is constant and equal to g,

$\mathrm{h}=\mathrm{ut}+\frac{1}{2}\mathrm{gt}{2 }^{ }$

$\Rightarrow 44.1=0+\frac{1}{2}(9.8){\mathrm{t}}^2$
$\therefore \mathrm{t}=3\mathrm{s}$

Let the initial speed of stone 2 be u, 

Both the stones strike the water simultaneously but stone 2 is thrown 1s later which means it takes 1s less as compared to stone 1 to cover the same distance.

Therefore, time taken by stone 2, 
$t_1=t- 1 =3-1 s = 2 s$
Applying the equation of motion,

$\mathrm{h}={\mathrm{ut}}_1+\frac{1}{2}{{\mathrm{gt}}_1}^2$

$\Rightarrow 44.1=\mathrm{u}\times 2+\frac{1}{2}\times 9.8\times {\left(2\right)}^2$

$\Rightarrow \mathrm{u}=12.25 \mathrm{m}/\mathrm{s}$