Answer
2.51 m
Work Step by Step
Let's take,
The velocity of the tile at the top of the window $=V$
Gravitational acceleration = g
Distance to the top of the window from the rooftop $=S$
Length of the window $=L$
Let's apply $V^{2}=u^{2}+2aS$ to the tile until it comes to the top of the window.
$\downarrow V^{2}=u^{2}+2aS$ ; Let's plug known values into this equation.
$V^{2}=0+2gS=>S=\frac{V^{2}}{2g}-(1)$
Let's apply $S=ut+\frac{1}{2}at^{2}$ to the motion of the tile from top ->bottom of the window
$\downarrow S=ut+\frac{1}{2}at^{2}$ ; Let's plug known values into this equation.
$1.6\space m =V(0.2\space s)+\frac{1}{2}\times 9.8\space m/s^{2}(0.2\space s)^{2}$
$V=\frac{3.2\space m-0.392\space m}{0.4s}=7.02\space m/s-(2)$
(2)=>(1)
$S=\frac{(7.02\space m/s)^{2}}{2\times9.8\space m/s^{2}}=2.51\space m$
