# Finding the Height of a Pot On the Y-Axis

## 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.

dx
Homework Helper
Gold Member
You can ignore the height of the pot.

Call the initial speed of the pot when it appears at the bottom of the window s. It will reach the top of the window after t = 0.25 seconds. Can you see why s must satisfy this equation:

st - (g/2)t2 = 2

g = 9.8 m/s2

t = 0.25 s

Solve for s.

Last edited:
Oh. I see. I just need to know the extra height added by the pot once it goes past the window, and not of the pot itself. I'm not 100% clear yet as to why t = 0.25 s, but I think it's because you're accounting for half of what is going on. Thanks, by the way.

dx
Homework Helper
Gold Member
The time 0.5 seconds is the total time that that the pot is in view, which is made up of both the time it is in view when it is going up and also the time it is in view when it is going down. One can show that these must be equal, so each time it will be in view for 0.25 seconds.

Once you find the initial speed s, you can calculate how far up it goes, and then subtract the height of the window from that to find out how far beyond the top of the window it goes.

Last edited: