## College Physics (4th Edition)

We can find the flowerpot's velocity at the top of the eighteenth-story window: $\Delta y = v_{0y}~t+\frac{1}{2}at^2$ $v_{0y}~t = \Delta y -\frac{1}{2}at^2$ $v_{0y} = \frac{\Delta y -\frac{1}{2}at^2}{t}$ $v_{0y} = \frac{1.5~m -(\frac{1}{2})(9.80~m/s^2)(0.044~s)^2}{0.044~s}$ $v_{0y} = 33.875~m/s$ We can let $v_f = 33.875~m/s$ and we can find the initial height $\Delta y$ above top of the eighteenth-story window: $v_f^2 = v_0^2+2a\Delta y$ $\Delta y = \frac{v_f^2-v_0^2}{2a}$ $\Delta y = \frac{(33.875~m/s)^2-0}{(2)(9.80~m/s^2)}$ $\Delta y = 58.55~m$ The flowerpot would need to have fallen from a height of 58.55 meters above the top of the eighteenth-story window. Since the twenty-fourth-story window is only 19 meters higher than the eighteenth-story window, the flowerpot could not have fallen with zero velocity from the twenty-fourth story window.