Question

A body falls from the top of the tower and during the last second of its fall it fall through 23mvfind height of tower.

Answer 1

Answer:

39.7 m

Explanation:

First, we conside only the last second of fall of the body. We can apply the following suvat equation:

$s=ut+\frac{1}{2}at^2$

where, taking downward as positive direction:

s = 23 m is the displacement of the body

t = 1 s is the time interval considered

$a=g=9.8 m/s^2$ is the acceleration

u is the velocity of the body at the beginning of that second

Solving for u, we find:

$ut=s-\frac{1}{2}at^2\\u=\frac{s}{t}-\frac{1}{2}at=\frac{23}{1}-\frac{1}{2}(9.8)(1)=18.1 m/s$

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:

$v^2-u^2 = 2as'$

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',

$s'=\frac{v^2-u^2}{2a}=\frac{18.1^2-0}{2(9.8)}=16.7 m$

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