Question

A drowsy cat is looking at a window from across the room, and sees a flowerpot that sail first up and then down past the window. The pot is in view for a total of 0.47 s, and the top-to-bottom height of the window is 2.14 m. How high above the window top does the flowerpot go?

Answer 1

How do you know the cat is drowsy? The question doesn't state whether or not the flowerpot leaves the frame of the window on the top end. If it passes the top of the window it could be up their for hours before it comes back down.

Answer 2

can't you figure out the velocity of the pot from the distance and the time? Wouldnt you half the time becasue it says a total of .47s?

Answer 3

I'm more concerned about the cat and why the flowerpot is sailing by the window. I think the cat got into some bad catnip.

Answer 4

matthewrlee - so funny, so unhelpful...

The pot is in view for a total of 0.47s, so its in view for half of that on the way up and half that on the way down. 0.47s/2 = 0.235s = t in the following equations.
The height of the window is 2.14 m. On the way up, the pot has an initial velocity of Vi as it passes the bottom of the window, and a velocity Vf when it reaches the top. Vf and Vi are related by \[v _{f} = v _{i} + at = v _{i} - g t\] We also know that the height of the window is equal to the average velocity times time:\[\Delta y = v _{avg} t\] and that \[v _{avg} = 1/2 (v _{f} + v _{i})\] so substituting for average velocity \[2\Delta y/t = v _{f} + v _{i}\] Now lets substitute for initial velocity: \[2\Delta y/t = v _{f} + g t + v _{f} = 2v _{f} + g t\] We know everything except final velocity, so we can solve for it: \[v _{f} = \Delta y /t - g t/2 = 2.14m/0.235s - 9.8 m/s ^{2} * 0.235s = 7.95m/s\] Now we can figure out how high it goes above the window top, using the fact that the velocity at the top of its flight (final velocity for this part of the problem) will be 0, and the velocity at the top of the window (initial velocity for this part of the problem) will be 7.95 m/s:\[v _{f}^{2} = v _{i}^{2} + 2 a \Delta y\] Substituting:\[0 = (7.95 m/s)^{2} + 2(-9.8m/s ^{2})(\Delta y)\] and solve for \[\Delta y = 3.22 m\]

Answer 5

lolcat