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Dennis_Churaev [7]

3 years ago

6

A drowsy cat spots a flower pot that sails first up and then down past an open window. The pot was in view for a total of 0.56 s

, and the top-to-bottom height of the window is 1.95 m. How high above the window top did the flowerpot go?

1 answer:

liberstina [14]3 years ago

4 0

Answer:

$h = 0.028 m$

Explanation:

As we know that

$d = \frac{v_2 + v_1}{2} t$

here we have

$1.95 = \frac{v_2 + v_1}{2}(0.56)$

$v_2 + v_1 = 6.96$

also we know

$v_2 - v_1 = at$

$v_2 - v_1 = (9.81)(0.56)$

$v_2 - v_1 = 5.49$

so we have

$v_2 = 6.23 m/s$

$v_1 = 0.74 m/s$

so the height above window is given as

$v_f^2 - v_i^2 = 2 a d$

$0.74^2 - 0 = 2(9.81)h$

$h = 0.028 m$

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