Assume that a free-falling object accelerates down at a constant rate. Assume, f

Question

Assume that a free-falling object accelerates down at a constant rate. Assume, further, that this rate, a is approximately 4 geshes/second^2. It TRULY does not matter if you have never heard of a gesh. A. One day, a cat sits near a window and watches the world outside. The length of the window is 3 geshes from top to bottom. There is a ledge above the window. The distance between the ledge and the top of the window is unknown. At some moment, all of a sudden, a flower pot falls off the ledge and plummets past the window. The cat notices that the flower pot takes 1/4 second to traverse the entire vertical length of the window. In geshes, how high above the top of the window is the ledge? B. (Part B is EXTRA CREDIT) Later, the cat observes the flower pot zoom up and then down past the window. The landlord, standing somewhere below the window, has attempted to throw the pot back onto the ledge. Evidently he missed. The flower pot, therefore, has gone up, reached some peak height and fallen down. In this scenario, therefore, a cat has observed the flower pot free-fall a full round-trip': up and then down. The pot has spent a total of 1 second in front of the window (up down) In geshes, how high above the top of the window was the peak height attained by the pot for this round trip?

Explanation / Answer

A)

a = 4 goshes/sec^2

L = 3 goshes

t = 1/4

let the initial velocity at the window is u

Using second equation of motion

3 = u * 0.25 + 0.5 * 4 * 0.25^2

solving for u

u = 11.5 gosh/sec

let the height is h

h = u^2/(2 * a)

h = 11.5^2/(2 * 4)

h = 16.5 gosh

the top of window is 16.5 gosh from ledge

b)

Now, total time = 1 s

time taken for going up = 1/2 = 0.50 s

3 = 0.50 * u + 0.50 * 4 * 0.50^2

u = 5 gosh/sec

peak height = 5^2/(2 * 4) = 3.13 gosh

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