Answer
The flowerpot fell from a window on the tenth floor.
Work Step by Step
We can find the flowerpot's velocity at the top of the window:
$\Delta y = v_{0y}~t+\frac{1}{2}at^2$
$v_{0y}~t = \Delta y -\frac{1}{2}at^2$
$v_{0y} = \frac{\Delta y -\frac{1}{2}at^2}{t}$
$v_{0y} = \frac{2.0~m -(\frac{1}{2})(9.80~m/s^2)(0.093~s)^2}{0.093~s}$
$v_{0y} = 21.0~m/s$
We can let $v_f = 21.0~m/s$ and we can find the initial height $\Delta y$ above top of the student's window:
$v_f^2 = v_0^2+2a\Delta y$
$\Delta y = \frac{v_f^2-v_0^2}{2a}$
$\Delta y = \frac{(21.0~m/s)^2-0}{(2)(9.80~m/s^2)}$
$\Delta y = 22.5~m$
Since the distance between each floor is $4.0~m$, the flowerpot fell from a window about 5.5 floors above the top of the student's fourth floor window. Therefore, the flowerpot fell from a window on the tenth floor.
