posted by Andrew on .
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 0.37 s, and the top-to-bottom height of the window is 2.20 m. How high above the window top did the flowerpot go?
S= ut +(1/2)at^2
S= (0)(0.215) + (0.5)(9.8)(0.215)^2
That's what I did and the site tells me I'm wrong. Please help.
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