Question

You drop a ball from a window located on an upper floor of a building. It strikes the ground with a speed $V$. You now repeat the drop, but your friend down on the ground throws another ball upward at the same speed $V$, releasing his ball at the same moment at which you drop yours from the window. At some location, the balls pass each other. This location is:

• A. At the halfway point between window and ground
• B. Above halfway point
• C. Below halfway point
• D. Cannot be determined

Answer 1

The correct option is B Above halfway point



For the ball that is dropped,
$V^2=2gH$ ($H$ is the height of the window above the ground)
Now,
$y=\frac{1}{2}g{t}^2$
$H-y=Vt-\frac{1}{2}g{t}^2\Rightarrow H=Vt$
So,
$y=\frac{1}{2}g{\left(\frac{H}{V}\right)}^2$
$=\frac{H^{2}g}{2}\times \frac{1}{2gH}$
$=\frac{H}{4}$