- #1

Juan42

- 5

- 0

## Homework Statement

A cat spots a flowerpot that sails first up and then down past an open window. The post is in view for a total of 0.50 seconds, and the top-to-bottom height of the window is 2.00 m. How high above the window top does the flower pot go? Answer: 2.34

## Homework Equations

y = (1/2)g*t^2 + Vo*t + yo

## The Attempt at a Solution

Let's see this is what I did. I tried to combine the known height of the window with the unknown height of the pot and used one of the above to do so.

hp = height of the pot

hw = height of the window

y = (1/2)g*t^2 + Vo*t + yo

hw + hp = (1/2)g*t^2 + Vo*t

2 + hp = (1/2)g*t^2 + 0*t

2 + hp = (1/2)(9.8)(0.50^2)

hp = (4.9)(0.25)

It's obvious that is wrong. I think I need to find it's initial velocity since it was already moving, but I'm not sure.