From what height was the pot dropped at?

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SUMMARY

The problem involves calculating the height from which a flowerpot was dropped, given that it takes 0.2 seconds to fall past a 1.9-meter-high window. Using the equation of motion, d = v1*t + (1/2)at^2, where a is 9.8 m/s², the initial distance calculation yields 0.196 meters. However, to find the total height, one must account for the distance fallen before reaching the window, leading to the conclusion that the pot was dropped from a height of 3.7 meters above the ground.

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Homework Statement


A falling flowerpot takes 0.2s to fall past a window which is 1.9m above ground. From what height was the pot dropped at?


Homework Equations


d = v1*t + (1/2)at^2


The Attempt at a Solution


I'm not sure what to do with the 1.9m above ground...

D = 0*0.2 + (1/2)(9.8m/s^2)(0.2)^2
D = 0.196

The book answer is 3.7m though, which is way off. I'm sure I have to use 1.9m to find the answer, but I'm at a loss right now. Help! Please and thank you!
 
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The flower pot has already been falling for some time before it passes the top of the window.

There are three things you know about this situation: acceleration, time to pass window, and vertical distance of window. What can this information tell you about how fast it was moving at the top of the window?
 

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