Answer
The stone was dropped from a height of 1.6 meters above the window.
Work Step by Step
Let y be 2.2 meters. Let $v_0$ be the speed when the stone is at the top of the window. Let the down direction be positive. Therefore,
$y = v_0t + \frac{1}{2}at^2$
$v_0 = \frac{y - \frac{1}{2}at^2}{t} = \frac{2.2 ~m - \frac{1}{2}(9.8 ~m/s^2)(0.31 ~s)^2}{0.31 ~s} = 5.6 ~m/s$
We can use the speed at the top of the window to find the height above the window where the stone was dropped. For this part of the problem, let v = 5.6 m/s.
$\Delta y = \frac{v^2 - v_0^2}{2a} = \frac{(5.6 ~m/s)^2 - 0}{(2)(9.8 ~m/s^2)} = 1.6 ~m$
The stone was dropped from a height of 1.6 meters above the window.
