Answer:
The pot was dropped from the fourth floor.
Step-by-step explanation:
The height and velocity of the pot is given by the following equations:
y = y0 + v0 · t + 1/2 · g · t²
v = v0 + g · t
Where:
y = height at time "t".
y0 = initial height.
v0 = initial velocity.
t = time.
g = acceleration due to gravity.
v = velocity at time "t".
Let´s place the origin of the frame of reference on the ground. We know that from the second floor (y = 8.000 m) the pot traveled 4.000 m in 0.2600 s. Then:
y = y0 + v0 · t + 1/2 · g · t²
4.000 m = 8.000 m + v0 · 0.2600 s - 1/2 · 9.8 m/s² · (0.2600 s)²
(4.000 m + 1/2 · 9.8 m/s² · (0.2600 s)² - 8.000 m) / 0.2600 s = v0
v0 = -14.11 m/s
This velocity is the velocity of the pot when it is at a height of 8.000 m (on the second floor).
Now let´s find how much time it took the pot to reach that velocity starting from rest (the pot was dropped, not thrown. Then v0 = 0).
v = v0 + g · t
-14.11 m/s = 0 - 9.8 m/s² · t
t = -14.11 m/s / - 9.8 m/s²
t = 1.440 s.
This is the time it takes the pot to reach the second floor (y = 8.000 m). So, let´s calculate y0 using the equation of height:
y = y0 + v0 · t + 1/2 · g · t² (v0 = 0)
8.000 m = y0 - 1/2 · 9.8 m/s² · (1.440 s)²
8.000 + 1/2 · 9.8 m/s² · (1.440 s)² = y0
y0 = 18.16 m
The pot was dropped from (18.16 m / 4 = 4.54 ) the fourth floor.