A boy sees a ball go up and then down through a window 2.45 m high. If the total time that ball is in sight in 1s, the height above the window the ball rises is approximately

$S = ut - \frac{1}{2}g{t}^2$
u = speed of the ball at bottom of the window 
Time duration for traveling window height one way = $\frac{1}{2}$ sec
$2.45 = u * \frac{1}{2} - \frac{1}{2} * 9.8 * {\left(\frac{1}{2}\right)}^2$
$2.45 = \frac{u}{2} - 1.225$

$u = 7.35 \raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$
$h = \frac{u^2}{2g} = 2.7565$
So, 2.7565−2.45=0.3065 m