Question

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 0.60 s, and the top-to-bottom height of the window is 2.10 m. How high above the window top did the flowerpot go?

Answer #1

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When it falls back from the maximum height ,

let its velocity on reaching the top of the window be u .

It travels a distance of 2.10 m down for a time of (0.60/2) s

We know,

s = u*t + 1/2*a*t^{2}

2.10 = u *(0.60/2) + 1/2*9.8*(0.60/2)^{2}

u = 5.53 m/s

Using V² = U² + 2*a*s

s = u²/ 2g

s = 5.53^{2} / 19.6

s = **1.56025 m.**

The flowerpot went up from the top of the window for a height of
**1.56025 m.**

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