## College Physics (4th Edition)

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.