Answer
The speed when it is 3.0 meters below the window is $~~11~m/s$
Work Step by Step
We can use conservation of energy to find the speed when it is 3.0 meters below the window:
$K_f+U_f = K_0+U_0$
$K_f = K_0+U_0-U_f$
$\frac{1}{2}mv_f^2 = \frac{1}{2}mv_0^2+mg(h_0-h_f)$
$v_f^2 = v_0^2+(2g)(3.0)$
$v_f = \sqrt{v_0^2+(2g)(3.0)}$
$v_f = \sqrt{(8.0~m/s)^2+(2)(9.8~m/s^2)(3.0~m)}$
$v_f = 11~m/s$
The speed when it is 3.0 meters below the window is $~~11~m/s$
