Question

Two tall buildings face each other and are $180\text{m}$ apart. With what velocity (in $\text{m/s}$) must a ball be thrown horizontally from a window $55\text{m}$ above the ground in one building, so that it enters a window $10.9\text{m}$ above the ground in the second building? (Take $g=9.8{\text{m/s}}^2$)

Answer 1

Let the horizontal velocity of the ball be $u$
Height between two windows $=h=44.1\text{m}$
Time taken by the ball to reach from one window to another $=t=\sqrt{\frac{2h}{g}}$
Now, $x=u\times t\Rightarrow 180=u\sqrt{\frac{2\times 44.1}{9.8}}$ or $180=u\times 3$ or $u=60{\text{m s}}^{-1}$

Hence, the option $\left(B\right)$ is correct.