given: \(u=25 m / s\)
time of flight \(T=5 s\)
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 \(h=\frac{u^{2}}{2 g} ; h=\frac{25^{2}}{2 \times 10}=\frac{625}{20}\) \(h=31.25 m\)
second, we can determine as well the time taken to reach this maximum height
using
$$
t=\frac{u}{g}=\frac{25}{10}=2.5 s
$$
If the time of flight i.e. the total duration in all is \(8 s\), and it took \(2.5 s\) to get to point \(B\), then it will take
\((8-2.5) s =5.5 s\) to descend from point \B TO C
Penultimately, we use this time \(t=5.5 s\) 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
$$
s=u t+\frac{1}{2} a t^{2}
$$
\(s_{y}\) or \(s=h+H\)
since we consider the motion from \(B\) to \(C\).
$$
\begin{aligned}
&h+H=0( t )+1 / 2 gt \\
&h+H=1 / 2 g t^{2} \\
&h+H=1 / 2 \times 10 \times 5.5^{2} \\
&h+H=151.25 m
\end{aligned}
$$
and lastly,
since \(h=31.25 m\)
\(\therefore H=151.25-31.25\)
\(H=120 m\)
Wow! This is a very tall building. Can you name any of such building around you?