Height-O Mania

Let the height of the roof above the top of the window be $h$ and the ball takes time $t$ to fall through the distance $h$ . The acceleration due to gravity $g=9.8$ $m/s^2$ . Therefore, $h=\cfrac{1}{2}gt^2, \quad \quad h+2=\cfrac{1}{2}g(t+0.1)^2$ Expanding the second equation: $h+2=\cfrac{1}{2}g(t^2+0.2t+0.01)$ $h+2=h + \cfrac{1}{2}g(0.2t+0.01)$ $2=0.049(20t+1)$ $\Rightarrow t=\left(\cfrac{2}{0.049}-1\right)\cfrac{1}{20}=1.99\approx 2s$ Therefore, $h=\cfrac{1}{2}gt^2=0.5\times 9.8\times 4=\boxed{19.6}$

Let the distance be X m and time to fall upto top of window Tsec.
The equations to reach top and bottom are:-
X = (1/2) * 9.8 * T^2 .......and ...........X + 2 = (1/2) * 9.8 *( T + 0.1 )^2
Hence (1/2) * 9.8 * T^2 + 2 = (1/2) * 9.8 *( T + 0.1 )^2
2 = (1/2) * 9.8 *( .2 * T + 0.01 ) ....<....> T = 1.99 sec.
X = (1/2) * 9.8 * (1.99)^2 = 19.6 m.

the ball travels the 2m distance in 0.1s

s = u t+1/2 g t^2

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

s = u(0.1) + 0.05

2 = u(0.1) + 0.05

2 - 0.05 = u(0.1)

1.95 = u(0.1)

1.95/0.1 = u

19.5 = u

now apply the kinametical equation

v^2 = u^2 + 2 a s .......................................... where v = 19.5 u = 0 a = 10m/s^S

and solve for s

s=19.6