Answer
The speed of the ball as it passes the top of the window is $~~22.0~m/s$
Work Step by Step
We can find the initial velocity $v_0$:
$y-y_0 = v_0~t+\frac{1}{2}at^2$
$v_0~t = y-y_0-\frac{1}{2}at^2$
$v_0 = \frac{y-y_0-\frac{1}{2}at^2}{t}$
$v_0 = \frac{12.2~m-36.6~m-\frac{1}{2}(-9.8~m/s^2)(2.00~s)^2}{2.00~s}$
$v_0 = -2.4~m/s$
We can find the velocity at the top of the window:
$v = v_0+at$
$v = -2.4~m/s+(-9.8~m/s^2)(2.00~s)$
$v = -22.0~m/s$
The speed of the ball as it passes the top of the window is $~~22.0~m/s$.
