A drowsy cat spots a flowerpot that sails first up and then down past an open window. The pot was in view for a total of 0.51 s, and the top-to-bottom height of the window is 2.25 m. How high above the window top did the flowerpot go?

Question

Physics

Muhammad Lester

31 December, 10:49

A drowsy cat spots a flowerpot that sails first up and then down past an open window. The pot was in view for a total of 0.51 s, and the top-to-bottom height of the window is 2.25 m. How high above the window top did the flowerpot go?

+2

Answers (1)

Know the Answer?

Teddy · Answer 1

The pot spends T = 0.185s going up and 0.185s going down past the window.

The average speed passing by the window is 2.20 m/0.185s = 11.89 m/s.

During passage, the pot increases speed by T*g = 0.185*9.81 = 1.815 m/s

The speed is therefore 12.80 m/s at the bottom of the window and 10.98 m/s at the top of the window.

The 10.98 m/s speed at the top of the window allows it to rise another 10.98^2 / (2g) = 6.15 m past the top of the window