A drowsy cat spots a flowerpot that sails up first and then down past an open wi

Question

A drowsy cat spots a flowerpot that sails up first and then down past an open window. The pot was in view for a total of 0.63 s, and the top-to-bottom height of the window is 2.50 m. How high above the window top did the flowerpot go? (In m)
Please explain variables and what equation was used. Thank you. A drowsy cat spots a flowerpot that sails up first and then down past an open window. The pot was in view for a total of 0.63 s, and the top-to-bottom height of the window is 2.50 m. How high above the window top did the flowerpot go? (In m)
Please explain variables and what equation was used. Thank you.
Please explain variables and what equation was used. Thank you.

Explanation / Answer

First we need to calculate the initial velocity

x = ut + 0.5at^2

2.50 = (u x 0.63) + 0.5 x (-9.8)(0.63)^2

u = 7.06 m/s

Using relation,

v^2 - u^2 = 2ay

0 - 7.06^2 = 2(-9.8)y

y = 2.54 m

So, 2.54 - 2.5 = 0.040 m above the window before coming back down.

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