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:

Let the time taken by 1st stone to reach water surface be t.

h = ut +1/2gt²

44.1 = 0 x t + 1/2x 9.8x t²

t= $\sqrt{\frac{2\times 44.1}{9.8}}$

  = 3 sec

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, the time taken by the second stone, t₁​=t−1   

                                                                           =3−1 s

                                                                           =2 s

Since acceleration is constant,

h = ut +1/2gt²

⇒44.1=u×2+$\frac{1}{2}$​×9.8×(2)²

⇒u=12.25 m/s