Finding the Height of a Pot On the Y-Axis

[Juan42]

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]

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.

 

[Juan42]

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]

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.

 

[Nickie]

First of all, great job attempting to use the equation for projectile motion to solve this problem! However, there are a few things that need to be addressed in your solution.

1. The equation you used, y = (1/2)g*t^2 + Vo*t + yo, is actually the equation for the vertical position of a projectile launched from the ground with an initial velocity, Vo. In this problem, the flowerpot is not launched from the ground, so we cannot use this equation directly.

2. In order to use the equation for projectile motion, we need to know the initial velocity of the flowerpot. This information is not given in the problem, so we cannot solve for the height using this method.

3. Instead, we can use the known height of the window and the time the flowerpot was in view to calculate the average velocity of the flowerpot. Then, we can use this average velocity in the equation y = Vo*t + yo to solve for the height of the flowerpot.

4. Using this method, we get the following equation:

hw = Vo*t + yo

Solving for the height of the flowerpot, we get:

hp = hw - yo = 2.00 m - 0 m = 2.00 m

Therefore, the height of the flowerpot is 2.00 m above the top of the window.

I hope this helps clarify the solution for you. Keep up the good work in your scientific endeavors!