Question

ball a is dropped from the top of a building. one second later ball b is dropped from the same point. as time progresses, the distance between them
multiple choice
need more information
decreases.
remains constant.
increases.

Explanation:

For a freely - falling object, the displacement formula is $y = \frac{1}{2}gt^{2}$. Let the time of fall of ball B be $t$, then the time of fall of ball A is $t + 1$. The distance of ball A from the top is $y_{A}=\frac{1}{2}g(t + 1)^{2}$, and the distance of ball B from the top is $y_{B}=\frac{1}{2}gt^{2}$. The distance between them $\Delta y=y_{A}-y_{B}=\frac{1}{2}g(t + 1)^{2}-\frac{1}{2}gt^{2}=\frac{1}{2}g(2t + 1)=gt+\frac{g}{2}$. As $t$ (time) increases, $\Delta y$ (the distance between the two balls) increases. So the distance between them increases as time progresses.

Answer:

D. increases.