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

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Discussion Overview

The discussion centers around a kinematics problem involving a flowerpot in free fall, specifically calculating the height above a window that the pot reaches after being spotted by a cat. The problem involves concepts of projectile motion and time of flight.

Discussion Character

  • Exploratory
  • Technical explanation
  • Homework-related

Main Points Raised

  • One participant seeks assistance in solving a kinematics problem involving a flowerpot's motion and expresses confusion over the variables involved.
  • Another participant suggests that the pot's trajectory is symmetrical, indicating that it takes half the total time to reach the peak height.
  • A participant questions whether it is always valid to assume that the time taken to ascend equals the time taken to descend in ideal free fall scenarios.
  • Another participant confirms that for an object thrown from the ground, the velocity at the same height during ascent and descent will be equal in magnitude but opposite in direction, supporting the symmetry of the motion.
  • A participant mentions using the falling body equation of motion to set up the problem, indicating a method to approach the solution.

Areas of Agreement / Disagreement

Participants generally agree on the symmetry of the motion in free fall, but the discussion does not reach a consensus on the specific approach to solving the problem or the handling of variables.

Contextual Notes

There are unresolved aspects regarding the application of kinematic equations and the assumptions made about the ideal conditions of free fall.

murphy
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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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