## Essential University Physics: Volume 1 (3rd Edition)

We know that the length of the window is given as $d=vt-\frac{1}{2}gt^2$ We plug in the known values to obtain: $-1.3=v(0.22)-\frac{1}{2}(9.8)(0.22)^2$ $v=-4.831\frac{m}{s}$ Hence the height above the window is: $h=\frac{v^2}{2g}$ We plug in the known values to obtain: $h=\frac{(4.831)^2}{2(9.8)}$ $h=1.2m$