A stone is dropped from the top of a building. When it crosses a point 5 m below the top, another stone starts to fall from a point 25m below the top. Both stones reach the bottom of building simultaneously. The height of the
building is

let the height of the tower is h

and time is taken to fall a distance 5 m by first stone is 1 sec

as

 $$\begin{gathered} s=\frac{1}{2}g{t}_1^2 \\ \Rightarrow 5=\frac{1}{2}\times 10\times t_1^2 \\ \Rightarrow t_1=1sec \end{gathered}$$

Now 

For the motion of the first stone

$h=\frac{1}{2}g{t}^2......\left(1\right)$

and For the motion of the second stone

$h-25=\frac{1}{2}g(t-1{)}^2......(2)$

From equation (1) and (2)

we get t=3 sec

So $h=\frac{1}{2}g{t}^2=5\times \left(3^2\right)=45 m$