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Homework Statement
An object falls from the top of a 103.6m high building. The vertical
distance between two consecutive windows A and B is 4.0m. The first window/storey is
also 4m from the top of the building. The object reaches window B 0.1s after it has reached window A.
Which storey is A ?
 Top of building, 103.6m above ground

 4m

= Window

 4m

= Window

4m

= Window
.
.
= Window A

4m,time to fall from A to B is 0.1s

= Window B



.
.
_ Ground
Homework Equations
d= 1/2gt^{2}
The Attempt at a Solution
The height of the building seems irrelevant (it is not a multiple of 4) except to indicate the object will be falling quickly enough to satisfy the conditions of the problem. What appears to be required is to relate the time interval (0.1s) to the distance interval 4n and 4n+1
The time for an object to fall a given distance from a start velocity of 0m.s will be
SQRT(2D/g)
The most promising approach appeared to be to relate the distance between the windows A and B expressed as 4n metres to the time interval and solve for n (the number of storeys from the top) so that:
time interval = time to reach A  time to reach B
Letting DB = distance from top to Window B, DA=distance from top to Window A
0.1 = SQRT(2DB/g)  SQRT(2DA/g)
Substituting DB = 4(n+1) metres, DA = 4n metres
0.1 = SQRT(2(4(n+1))/g)  SQRT(2(4n)/g)
Sadly I'm in error because the n's cancel!
What have I done wrong?