Physics Question- Free Fall?

Question

a roof tile falls from rest from the top of a building. an observer inside the building notices it takes 0.20 seconds for the tile to pass her window. whose height ius 1.6m. how far above the top of this window is the roof?

If anyone could help me, thanks alot! I am really confused what to do

Answer 1

dH=H2-H1
dt=t2-t1
H=0.5 g t^2

dh - height of the window
H2- distance from the roof to the botom of the window
H1- distance from the roof to the top of the window
dt -time the observes sees the tile passing by the window

dh=.5g(t2^2 - t1^2)
or
dh=.5 g(t2 - t1)(t2+t1)

Now can you finish by yourself?
No? we have one too many unknowns?
Okay...
Let's see the distance the tile moves across the window is
dh=V0dt + 0.5g (dt)^2
so V0=(dh-0.5g (dt)^2)/dt
and
V0=gt1 so t1 finally is
t1=Vo/g
and
t1=(dh-0.5g (dt)^2)/(g dt )

So since
H1=0.5g t1^2
H1=0.5g [(dh-0.5g (dt)^2)/(g dt )]^2
H1=0.5[(dh-0.5g (dt)^2)/(dt )]^2/g

Answer 2

2.511 m.
First we solve for v1 (velocity at top of window) then for s1 (fall distance to top of window). t1 is the time to fall to s1, t is time to fall past (from top to bottom of) window (=0.2), v is the mean velocity while falling past window, a is acceleration = g = 9.81 m/s^2.
v = 1.6/0.2 = 8 m/s
v1 = v - at/2 = 7.019 m/s
s1 = at1^2/2 = v1*t1/2 = v1^2/(2a) = 2.511 m