Question

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?

Answer 1

about 2.4m above the window

Answer 2

`h = 0.028 m`

Step-by-step explanation:

As we know that

`d = (v_2 + v_1)/(2) t`

here we have

`1.95 = (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`