Urgent physics question

A stone is dropped into a river from a bridge 44.6 m above the water. Another stone is thrown vertically down 1.33 s after the first is dropped. Both stones strike the water at the same time. What is the initial speed of the second stone?

any help will be much appreciated

for the stone that was dropped with no initial velocity ( free fall motion) :
---> h = 1/2 g t^2 ----> 44.6 = 0.5 x 10 t^2 ---> t = √44.6/5 = 2.99 s
for the stone that was thrown vertically down will reach the water at t' = 2.99 - 1.33 = 1.66 s
---> h = vo.t + 0.5 g t^2
44.6 = vo.1.66 + 5(1.66)^2
44.6 = vo.1.66 + 13.78 ----> vo = (44.6 - 13.78)/1.66 = 18.57 m/s

Put both stones into their own equation. Use the equation y=yo+v(t)+(1/2)a(t^2)

so you get 44.6=4.8(t^2) Solve for t
then
t-1.33=the t for the below equation
44.6= v(t)+4.8(t^2)

This leave only v left. answered!