Gravity Height of the roof

Let the time for the ball to reach the top of the window be $t_1$ . Since the distance dropped is given by $s(t) = \frac {1}{2} g t^2$ then, we have (assuming $g = 9.8ms^{-2}$ ):

$s(t_1+0.1)-s(t_1) = \frac {1}{2} g [(t_1+0.1)^2-t_1^2]$

$2 = 4.9 (t_1^2+0.2t_1+0.01-t_1^2)$

$\Rightarrow 0.98t_1 = 2 - 0.049 \quad \Rightarrow t_1 = 1.99 s$

Therefore, the height of the roof above the top of the window:

$s(t_1) = 4.9 t_1^2 = 4.9\times 1.99^2 = \boxed{19.4} m$

Time taken to cross the window of height 2 m is 0.1 sec so

s = ut + 1/2 gt^2

2 = u x 0.1 + 5(0.1)^2

2 = u x 0.1 + 0.05

19.5 = u

now substitute the initial velocity in eq

s = (v^2 - u^2)/2