Let the height of the building be $H$.

• Stone 1 is dropped from the top.
• When Stone 1 has fallen $5\text{ m}$, Stone 2 is released from a point $25\text{ m}$ below the top.

Find the delay before Stone 2 is released.

Stone 1 starts from rest and falls $5\text{ m}$: $5=\frac{1}{2}gt_1^2$ $t_1=\sqrt{\frac{10}{g}}$

This is the time delay between the release of the two stones.

Let $T$ be the total time taken by Stone 1 to reach the ground.

Then Stone 2 falls for time: $T-t_1$

Write the distance equations.

For Stone 1, falling from the top through height $H$: $H=\frac{1}{2}gT^2$

For Stone 2, released from $25\text{ m}$ below the top, so it falls distance: $H-25$ Hence, $H-25=\frac{1}{2}g(T-t_1)^2$

Subtract the two equations.

$H-(H-25)=\frac{1}{2}g\left[T^2-(T-t_1)^2\right]$ $25=\frac{1}{2}g\left[2Tt_1-t_1^2\right]$

Now use: $t_1^2=\frac{10}{g}$

So, $25=\frac{1}{2}g\left(2Tt_1-\frac{10}{g}\right)$ $25=gTt_1-5$ $gTt_1=30$

Substitute $t_1=\sqrt{10/g}$.

$gT\sqrt{\frac{10}{g}}=30$ $T\sqrt{10g}=30$

Taking $g=10\text{ m/s}^2$, $T\sqrt{100}=30$ $10T=30$ $T=3\text{ s}$

Now find the height $H$.

$H=\frac{1}{2}gT^2$ $H=\frac{1}{2}\cdot 10\cdot 3^2$ $H=45\text{ m}$

Check options.

The correct option is: $\boxed{\text{C: }45\text{ m}}$