Question

Two tall buildings are $40\text{m}$ apart. With what speed a ball must be thrown horizontally from a window $145\text{m}$ above the ground in one building, so that it will enter a window $22.5\text{m}$ above the ground in the other ?
(Take $g=10m/s^2$)

• A. $5{\text{ms}}^{-1}$
• B. $8{\text{ms}}^{-1}$
• C. $10{\text{ms}}^{-1}$
• D. $16{\text{ms}}^{-1}$

Answer 1

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


Let the hprizontal velocity of ball be $v$
Vertical distance between the two windows $=h=145-22.5=122.5\text{m}$
Time taken by the ball to reach from one window to another $=t=\sqrt{\frac{2h}{g}}$
Therefore, horizontal distance covered
$x=v_{x}t$
$\Rightarrow 40=v\sqrt{\frac{2\times 122.5}{10}}$
$\Rightarrow 40\approx v\times 5$
$\Rightarrow v\approx 8{\text{ms}}^{-1}$