$A$ stone is dropped from the edge of a roof. It passes a window $2 \, m$ high in $0.1 \, s$. How far is the roof above the top of the window?

(19.42 M) Let the distance between the top of the window and the roof be $S$. For the journey across the window,i.e.,from $B$ to $C$:
Let the velocity at $B = u \, m/s$.
Distance travelled,$S = 2 \, m$.
Time taken,$t = 0.1 \, s$.
Acceleration,$a = g = 9.8 \, m/s^2$.
Using the equation $S = ut + \frac{1}{2}at^2$,we have:
$2 = u(0.1) + \frac{1}{2}(9.8)(0.1)^2$
$2 = 0.1u + 0.049$
$0.1u = 1.951$
$u = 19.51 \, m/s$.
Therefore,the velocity of the stone at the top of the window is $19.51 \, m/s$.
Now,for the free fall of the stone from the roof top to the top of the window,i.e.,from point $A$ to $B$:
Initial velocity $u_i = 0 \, m/s$.
Final velocity $v = 19.51 \, m/s$.
Distance covered $S = ?$.
Acceleration due to gravity $g = 9.8 \, m/s^2$.
Using $v^2 - u_i^2 = 2gS$,we have:
$(19.51)^2 - 0^2 = 2 \times 9.8 \times S$
$S = \frac{(19.51)^2}{19.6} \approx 19.42 \, m$.