's' increases at all times that Rock X falls since Rock X falls with a greater velocity than Rock Y.
<h3>What is a frictional force?</h3>
The friction force is the opposite force generated when two parts slide over each other if the air is passing over the rocks the frictional force would be the drag force.
Given: Frictional forces are considered to be negligible.
A few seconds after Rock X is released from rest, Rock Y is also released. Consider moving downward to be constructive. Both rocks are launched from rest, thus their initial velocities are equal to zero (u = 0).
We know that the velocity is given by,
v² = u² + 2as
Rock X is released first, followed by Rock Y, and because Rock X travels faster than Rock Y, it travels farther. This is due to the fact that whenever Rock X has already acquired some velocity (t).
Its velocity (v) is higher, increasing the spacing (s). In conclusion, because Rock X falls faster than Rock Y,'s' rises each time Rock X hits the ground.
The complete question is given below:-
Rock X is released from rest at the top of a cliff that is on Earth. A short time later, Rock Y is released from rest from the same location as Rock X. Both rocks fall for several seconds before landing on the ground directly below the cliff. Frictional forces are considered to be negligible. After Rock Y is released from rest several seconds after Rock X is released from rest, what happens to the separation distance S between the rocks as they fall but before they reach the ground, and why? Take the positive direction to be downward.
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