given: $u=25m/s$
time of flight $T=5s$
we are required to find the height of the building $H=$ ? (See diagram below)
to ease our task, we present the diagram below to describe the problem at hand.irst we determine the maximum height $h$ of projection (from point $A$ to $B$ )
using $h=\frac{u^2}{2g};h=\frac{{25}^2}{2\times 10}=\frac{625}{20}$ $h=31.25m$
second, we can determine as well the time taken to reach this maximum height
using
$$t=\frac{u}{g}=\frac{25}{10}=2.5s$$
If the time of flight i.e. the total duration in all is $8s$, and it took $2.5s$ to get to point $B$, then it will take
$(8-2.5)s=5.5s$ to descend from point \B TO C
Penultimately, we use this time $t=5.5s$ to calculate the total height which is $H+h$ (see diagram)
If we consider the motion from $B$ to $C$, we see that at point $B$, our initial velocity $u$ is zero
using
$$s=ut+\frac{1}{2}a{t}^2$$
$s_y$ or $s=h+H$
since we consider the motion from $B$ to $C$.
$$\begin{array}{rl}& h+H=0(t)+1/2gt\\ & h+H=1/2g{t}^2\\ & h+H=1/2\times 10\times {5.5}^2\\ & h+H=151.25m\end{array}$$
and lastly,
since $h=31.25m$
$\therefore H=151.25-31.25$
$H=120m$
Wow! This is a very tall building. Can you name any of such building around you?