## University Physics with Modern Physics (14th Edition)

Let $h$ be the vertical length of the window. Let $v_0$ be the speed when the pot reaches the top of the window. $h = v_0t + \frac{1}{2}at^2$ $v_0 = \frac{h}{t} - \frac{1}{2}at$ $v_0 = \frac{1.90~m}{0.380~s} - \frac{1}{2}(9.80~m/s^2)(0.380~s)$ $v_0 = 3.138~m/s$ We can use this velocity as $v$ for the next step of the solution. Let $y$ be the height from the top of the window to the windowsill where the flowerpot fell. $y = \frac{v^2-v_0^2}{2g} = \frac{(3.138~m/s)^2-0}{(2)(9.80~m/s^2)}$ $y = 0.502~m$ The windowsill is 0.502 meters above the top of the window.