Help with Problem (Complicated Kinematics)

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Homework Help Overview

The discussion revolves around a kinematics problem involving a flowerpot that moves past a window, with participants exploring the dynamics of its motion, including the effects of gravity and varying velocities.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss calculating the average speed of the flowerpot and consider the relationship between its velocities during upward and downward motion. Some suggest drawing diagrams to visualize the problem, while others express confusion about the initial approach and the implications of gravity on the motion.

Discussion Status

The discussion is active, with various approaches being suggested, including breaking the problem into steps and questioning the assumptions made about the scenario. Some participants have offered hints and guidance, but no consensus has been reached on a specific method to solve the problem.

Contextual Notes

Participants have raised questions about the realism of the scenario presented, including the motivations of the cat and the feasibility of measuring time in this context, which may affect their interpretations of the problem.

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Help with Problem please (Complicated Kinematics)

A drowsy cat spots a flowerpot that sails first up and then down past an open window. The pot was in view for a total of 0.35 s, and the top-to-bottom height of the window is 1.95 m. How high above the window top did the flowerpot go?

-At first I tried finding the velocity by m/s. Then later when I thought about it, the velocity is not constant in this problem. Now I have no clue what to do, even though I know 9.8 m/s^2 as gravity and the velocity in the distances of the window.
 
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You'll need to attack this problem in steps. There are several ways to go. Try figuring out the speed of the flowerpot at various points. Hint: Start by finding the average speed of the flowerpot as it passes by the window.
 
Think about the problem. Will the velocities up and down be the same? If so, then you can calc the initial and final velocities of the pot going past the window, and the rest is easy peasy. Try drawing a diagram also.
 
The up and down speeds of the pot is the same at the same height above the ground, they are just in opposite directions. This can be seen from the equation

v^2 = u^2 - 2gy

which do not distinguish between the up and down motions. This means that the up and down times to cross the window will be the same when one considers

v = u - gt

for the separate cases of the up and down motion past the window. For instance say the velocity at the bottom is va and the top vb, then for the going up case (taking up as positive) we have

v_b = v_a - gt

and for the going down case (taking down as positive) we have

v_a = v_b + gt

From this info you can calculate va from

y = v_at - 4.9t^2

But I must say I see several logical errors in the problem:
1. Why would someone want to throw a flower pot upwards? He would just have to catch it again on its way down or jump out of the way.
2. How or why would the cat measure the time for the pot to cross the window?
3. Who taught him to add?
4. How did he communicate his results to the physicist?
5. If the cat did not do it, why is he mentioned in the problem anyway?
 
Last edited:

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