A cat spots a flower pot through the window

Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
7 replies · 3K views
Akash47
Messages
53
Reaction score
5

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 0.50 s, and the top-to-bottom height of the window is 2.00 m. How high above the window top does the flowerpot go?

2. Relevant equation
Not sure but the theory of refraction of light may work.

The Attempt at a Solution


I didn't understand the question at all.That's why I want to know what kind of question this is.And also what is it's solution process?
 
Physics news on Phys.org
Shorn of hyperbole and ornamentation, the question seems to be...

"A flowerpot is thrown upwards, passing by a 2m window both on the way up and back down. The total time taken passing the window (twice) is 0.5s . How high above the top of the window is the top of the object's travel ?"

This is a question of projectile motion; neither optics, nor animal behaviour.
 
Last edited:
  • Like
Likes   Reactions: DrClaude
But there is no mention of throwing the flower pot.
 
One has to imagine that some unseen person perhaps another cat has thrown the flower pot upward or that the flower pot was on a bench and a cat jumped down on the bench catapulting the flower pot upward. There are innumerable scenarios where the flower pot can be launched into the air. However none of that matters once the pot is in motion. We also have to assume that this event occurred on the Earth as we know of no other planet where there are cats and flower pots.

Following the symmetry of the situation, if the pot goes up then the pot comes down. It will traverse the same window length in the same amount of time.

Given that you can you conclude?
 
Akash47 said:

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 0.50 s, and the top-to-bottom height of the window is 2.00 m. How high above the window top does the flowerpot go?

2. Relevant equation
Not sure but the theory of refraction of light may work.
I always use the "kinematics equations" when I run across problems like this.
Here are the ones I keep handy:

kinematics equations
d = vi * t + 1/2 * a * t^2
vf^2 = vi^2 + 2 * a * d
vf = vi + a * t
d = (vi + vf) / 2 * t

d: distance
a: acceleration
t: time
vi: initial velocity
vf: final velocity​

Put them in your toolbox!

The Attempt at a Solution


I didn't understand the question at all.That's why I want to know what kind of question this is.And also what is it's solution process?

I think others have adequately figured out the problem.
My solution process started with identifying the known variables: acceleration, distance, and time.

Then I rearranged the kinematics equations to put all of the unknown variables on one side.
Did a bunch of maths, and solved the problem.

jedishrfu said:
One has to imagine that some unseen person perhaps another cat has thrown the flower pot upward or that the flower pot was on a bench and a cat jumped down on the bench catapulting the flower pot upward.

In my scenario, a cat on the x floor of an apartment building knocked the flower pot off a banister.
The flower pot interacted with a trampoline at the bottom.
The cat in question, noticed the flower pot on its subsequent upward and downward travels.

ps. Fun problem!