A steel ball is dropped from a building's roof and passes a window of lenght $10\text{m}$, taking $1\text{s}$ to fall from the top of the window to the bottom of the window. It then falls onto a sidewalk and bounces back past the window, moving from bottom to top in $1\text{s}$. Assume that the upward flight is an exact reverse of the fall. The time the ball spends below the bottom of the window is $2\text{s}$. How tall is the building?

The correct option is D

31.25 m

$BC$ is the window. When the ball falls from $B$ to $C$.

$s=-10\text{m}$,
$a=-10{\text{m/s}}^2$,
$t=1\text{s}$

Using
$s=ut+\frac{1}{2}a{t}^2$

$\Rightarrow -10=u\left(1\right)+\frac{1}{2}(-10)(1{)}^2$

$\Rightarrow -10=u-5$

$\Rightarrow u=-5m/s$

$\therefore$Velocity at $B=-5\text{m/s}$

Velocity at $C=u+at=-5-10\left(1\right)=-15\text{m/s}$

Time for $CD$ is $1\text{s}$

$\therefore CD=s$ $=ut+\frac{1}{2}a{t}^2$

$\Rightarrow -15\left(1\right)+\frac{1}{2}(-10)(1{)}^2=-15-5=20\text{m}$

$\therefore CD=20\text{m}$

To find $AB$

Using,
$v^2=u^2+2as$

$\Rightarrow (-5{)}^2=0+2(-10\left)\right(s)$

$\Rightarrow 25=0-20s$

$\Rightarrow s=\frac{-25}{20}=\frac{-5}{4}=-1.25\text{m}$ $\therefore AB=1.25\text{m}$

$\therefore$ Total height $=AB+BC+CD=1.25+10+20=31.25\text{m}$