Calculating the Height of a Flowerpot in Free Fall: A Kinematics Problem

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The discussion centers around solving a kinematics problem involving a flowerpot in free fall, observed by a cat. The pot is visible for 0.50 seconds as it travels 2.00 meters past a window, and the goal is to determine how high it ascends above the window. It is established that the pot's trajectory is symmetrical, taking 0.25 seconds to rise and 0.25 seconds to fall, confirming that the time to ascend equals the time to descend. Participants emphasize the importance of using the falling body equations to simplify calculations. The final answer for the height above the window is determined to be 2.34 meters.
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The Drowsy Cat - Kinematics!

Can anybody help me get started? I have been working on this problem forever...
A drowsy cat spots a flowerpot that sails first up and then down past an open window. The pot is in view for a total of 0.50s and the top to bottom height of the window is 2.00m. How high above the window does the flowerpot go? The answer is 2.34m but I don't know how to get it, I know that it is a kinematics free fall question but I keep getting too many variables. If anyone knows how to go about answering this, that would be awesome! Thank you!
 
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The pot's arc will be symmetrical, that means it will take exactly half the time (0.25 seconds) to travel upwards the 2m. (The second half of the pot's trajectory is irrelevant.) Does that simplify things?
 
Yes! Thank you very much. So in these problems, then, can you always assume that the time to go up = the time to come back down? (assuming an ideal free fall situation)
 
For an object thrown from the ground, its velocity on the way down will be equal in magnitude but opposite in direction then at the same height on the way up. So yes it will take the same amount of time to pass the window going going up as going down.

If you set up the falling body equation of motion with t=0 and x=0 at the bottom of the window it works out nicly.
 
My stars & garters - I helped someone! :-)
 

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