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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?

Respuesta :

Answer:

[tex]h = 0.028 m[/tex]

Explanation:

As we know that

[tex]d = \frac{v_2 + v_1}{2} t[/tex]

here we have

[tex]1.95 = \frac{v_2 + v_1}{2}(0.56)[/tex]

[tex]v_2 + v_1 = 6.96[/tex]

also we know

[tex]v_2 - v_1 = at[/tex]

[tex]v_2 - v_1 = (9.81)(0.56)[/tex]

[tex]v_2 - v_1 = 5.49[/tex]

so we have

[tex]v_2 = 6.23 m/s[/tex]

[tex]v_1 = 0.74 m/s[/tex]

so the height above window is given as

[tex]v_f^2 - v_i^2 = 2 a d[/tex]

[tex]0.74^2 - 0 = 2(9.81)h[/tex]

[tex]h = 0.028 m[/tex]

wolfiaat

Answer:

about 2.4m above the window

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