Question

A person throws ball with a velocity $‘v^{\prime }$ from top of a building in vertically upward direction. The ball reaches the ground with a speed of ${}^{\prime }3v^{\prime }$. The height of the building is:

• A. $\frac{4v^2}{g}$
• B. $\frac{3v^2}{g}$
• C. $\frac{6v^2}{g}$
• D. $\frac{9v^2}{g}$

Answer 1

The correct option is A $\frac{4v^2}{g}$

Let $H$ be the height of the building, then the total displacement of the ball is $H$. From third equation of motion
$v^2=u^2+2aS$
In this case, considering the downward direction as positive
$9v^2=v^2+2gH$
$8v^2=2gH$
$H=4v^2/g$
Therefore, option A is correct.