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

Given that

A stone is dropped from height h = 44.1

Let time taken = t, u = 0

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

$44.1=0+\frac{1}{2}{\text{gt}}^2\cdots \cdots \left(1\right)$

A second stone is thrown vertically down with initial velocity = v

u = v

Ti me taken to reach the water ${\text{t}}_{\text{1}}=\text{t}-1\text{s}$

Height descended h = 44.1m

$\text{s}=\text{vt}+\frac{1}{2}a{\text{t}}^{\text{2}}$

$44.1=\text{v}\left(\text{t}-1\right)+\frac{1}{2}\text{g}{\left(\text{t}-\text{1}\right)}^2\cdots \cdots \left(2\right)$

From (1)

$\frac{1}{2}{\text{gt}}^2=44.1$

${\text{gt}}^2=88.2$

${\text{t}}^2=\frac{88.2}{9.8}=9\Rightarrow \text{t}=\text{3s}$

Substitute t = 3 s in (2)

$\text{44}\text{.1=v}\left(3-1\right)+\frac{1}{2}\text{g}{\left(3-1\right)}^2$

$44.1=2\text{v}+\frac{1}{2}\left(9.8\right){\left(2\right)}^2$

$2\text{v}=44.1-19.6$

$\text{v}=\frac{24.5}{2}=12.25{\text{ms}}^{-1}$