A drowsy cat is looking at a window from across the room, and sees a flowerpot that sail first up and then down past the window. The pot is in view for a total of 0.47 s, and the top-to-bottom height of the window is 2.14 m. How high above the window top does the flowerpot go?

Question

Makenzie Chan

Mathematics

28 December, 09:13

A drowsy cat is looking at a window from across the room, and sees a flowerpot that sail first up and then down past the window. The pot is in view for a total of 0.47 s, and the top-to-bottom height of the window is 2.14 m. How high above the window top does the flowerpot go?

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Answers (1)

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Imani Gallagher · Answer 1

Time while going up = time while going down.

One-way time = 0.47 / 2 = 0.235 s

While going up,

s = ut + 1/2 at²; where u is the initial velocity upwards and a is the acceleration due to gravity downwards.

2.14 = u (0.235) + 1/2 x - 9.81 x (0.235) ²

u = 10.2 m/s

Max height attained can be obtained by:

2as = v² - u²; where v is 0

s = - (10.2) ² / - 9.81 x 2

s = 5.30 m

The height above the window is:

5.30 - 2.14

= 3.16 m