Answer:
39.7 m
Explanation:
First, we conside only the last second of fall of the body. We can apply the following suvat equation:
[tex]s=ut+\frac{1}{2}at^2[/tex]
where, taking downward as positive direction:
s = 23 m is the displacement of the body
t = 1 s is the time interval considered
[tex]a=g=9.8 m/s^2[/tex] is the acceleration
u is the velocity of the body at the beginning of that second
Solving for u, we find:
[tex]ut=s-\frac{1}{2}at^2\\u=\frac{s}{t}-\frac{1}{2}at=\frac{23}{1}-\frac{1}{2}(9.8)(1)=18.1 m/s[/tex]
Now we can call this velocity that we found v,
v = 18 m/s
And we can now consider the first part of the fall, where we can apply the following suvat equation:
[tex]v^2-u^2 = 2as'[/tex]
where
v = 18 m/s
u = 0 (the body falls from rest)
s' is the displacement of the body before the last second
Solving for s',
[tex]s'=\frac{v^2-u^2}{2a}=\frac{18.1^2-0}{2(9.8)}=16.7 m[/tex]
Therefore, the total heigth of the building is the sum of s and s':
h = s + s' = 23 m + 16.7 m = 39.7 m
