# 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

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: $v_1^2-0=2gy$.