- #1

QuantumCurt

Education Advisor

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## Homework Statement

A cat spots a flowerpot that sails first up and then down past an open window. The pot is in view for a total of .50s and the top-to-bottom height of the window is 2.00m. How high above the window does the flower pot go?

I was trying to help one of my friends with this problem, and apparently I'm not setting it up right. I can't figure out what I'm doing wrong.

## The Attempt at a Solution

I figured the velocity would be 8.0 m/s, because it's in view for a total of .50 seconds, between the two passes that it is making by the window.

So, one pass is 2.00m/.25 s=8 m/s

Then I used a kinematic to solve for displacement.

[tex]v^2=v_{0}^2+2a\Delta{x}[/tex]

Now I let the final velocity be 0 at the top of the flight path, and I substitute g in for a, with downward being the negative direction. Then solve for delta x, and substitute numerically.

[tex]\Delta{x}=\frac{-v^2_{0}}{-2g}[/tex]

[tex]\Delta{x}=\frac{-64}{-2(9.80)}[/tex]

Which gives me [itex]\Delta{x}=3.3 m[/itex] abpve the window.

Her book apparently says the answer is 2.34 m. Where did I go wrong?

I feel like the velocity calculation is kinda questionable here. I can't think of how else to go about it though. I tried a couple different approaches, and came up with answers that were much farther off.