Two tall buildings are $40\text{m}$ apart. With what speed must a ball be thrown horizontally from a window $145\text{m}$ above the ground in one building, so that it will enter a window $22.5\text{m}$ from the ground in the other ?

The correct option is B $8{\text{ms}}^{-1}$

Vertical distance between the two windows $=h=122.5m$
$\therefore$ Time taken by the ball to reach from one window to the other $=t=\sqrt{\frac{2h}{g}}$

Let the ball be projected be with a horizontal velocity $v$.
Now, horizontal distance travelled $x=v_{x}t$
$\Rightarrow 40=v\sqrt{\frac{2\times 122.5}{10}}\Rightarrow 40\approx v\times 5\Rightarrow v\approx 8m{s}^{-1}$