Answer:
Ok, first, suppose that a given object is dropped or thrown down.
Then the acceleration of the object is the gravitational acceleration, given by:
a(t) = -9.8m/s^2
The velocity of the object can be integrated from that, it gives:
v(t) = (-9.8m/s^2)*t + V0
where V0 is the initial speed of the object, in the case that it is dropped, V0 is equal to 0m/s
The position equation can be found integrating again:
p(t) = (1/2)*(-9.8m/s^2)*t^2 + V0*t + p0
Where p0 is the initial position.
Now let's see our problem.
We know that in 8 frames, the flowerpot falls 0.84 of the height of the window, which is 1.27m
Then the distance that it falls is:
D = 0.84*1.27m = 1.07m
Now we also know that the camera captures at 30 fps
Then we have the relation:
30 frames = 1 second
1 = (1 second)/(30 frames)
With this, we can rewrite:
8 frames = 8 frames* (1 second)/(30 frames) = 0.267 seconds.
So now we know the distance that the pot fall, and the time in which it did fall.
Remember that the position equation for a falling object, is:
p(t) = (1/2)*(-9.8m/s^2)*t^2 + V0*t + p0
let's define p0 = 0m
Then the position equation of the pot is given by:
p(t) = (1/2)*(-9.8m/s^2)*t^2 + V0*t
The distance that the pot would travel between t = 0s and t = t' is:
distance = p(t') - p(0s)
So, knowing that between t= 0s and t = 0.267s, the pot did travel -1.07m (the negative sign is because it traveled downwards), we can find the initial speed of the pot.
-1.07m = p(0.267s) - p(0s)
-1.07m = (1/2)*(-9.8m/s^2)*(0.267s)^2 + V0*0.267s
[-1.07m + (1/2)*(9.8m/s^2)*(0.267s)^2]/0.267s = V0 = -2.7 m/s
This means that the pot was thrown downwards with an initial speed of 2.7 m/s
So we can conclude that the pot did not fall down on its own, someone threw it intentionally downwards.
This is the only thing that we can conclude with the given information.
