Question

You drop a rock from a bridge to the river below. When the rock has fallen 4.0 meters you drop a second rock. As the rocks continue their free fall, does their separation increase, decrease or stay the same?

Answer 1

Their separation will increase. Why?

The distance covered by the first rock is h₁=4 m, you drop another one (with zero distance covered h₂=0), the initial separation is

$$\Delta h=|h_2-h_1|=|0-4|=4\text{ m}.$$

The time it took for the first rock to cover 4 m is

$$t_1=\sqrt{\frac{2h_1}{g}}=\sqrt{\frac{2\cdot4}{9.8}}=0.9\text{ s}.$$

The initial velocity of the second rock is of course 0. Let us calculate the distance the second rock covered after 2 seconds:

$$H_2=\frac{gt^2}{2}=19.6\text{ m}.$$

For the first rock, it will be 2.9 s. The distance of the first rock after 2.9 seconds is

$$H_1=\frac{9.8\cdot2.9^2}{2}=41.2\text{ m}.$$

The separation between the rocks 2 seconds after the second rock was thrown is

$$\Delta H=|H_2-H_1|=21.6\text{ m}.$$

As we see, the separation increases as both rocks continue to fall.