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A flowerpot falls from the ledge of an apart- ment building.

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A flowerpot falls from the ledge of an apartment building. A person in an apartment below, coincidentally holding a stopwatch, notices that it takes 0.9 s for the pot to fall past his window, which is 11 m high.

How far above the top of the window is the ledge from which the pot fell? The acceleration of gravity is 9.81 m/s2 .

Answer in units of m

asked Sep 15, 2014 in AP Physics C by anonymous

1 Answer

0 votes
For the pot to fall through 11 m height (h) with initial velocity (||v_1||) and time duration of 0.9 (||t_1||), we can find the initial velocity ||h=v_1t_1+\frac 12 gt_1^2||. Here we define downward as the positive direction.

With ||v_1|| found, one can figure out how much distance above the top of the window (y) the pot fell from: ||v_1^2-0=2gy||.
answered Sep 15, 2014 by PhysicsDude (21,750 points)
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