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.

Since rock X is first released followed by Y, then X has a greater speed than Y therefore the distance covered by X is longer. This is because despite 0.5gΔt² being same for both rocks at any time Δt but rock X having already attained some velocity, its ∪Δt is more hence the separation S increases. Conclusively, S increases at all times that rock X falls since rock X falls with a greater velocity than rock Y

ahirohit963 ahirohit963 · Answer 2 · 2021-10-25T08:27:37+02:00

's' increases at all times that Rock X falls since Rock X falls with a greater velocity than Rock Y.

Given :

Frictional forces are considered to be negligible.

Rock Y is released from rest several seconds after Rock X is released from rest.

Take downward direction to be positive.

Both Rocks release from rest therefore initial velocity of both the Rocks is zero that is u = 0.

Solution :

We know that the velocity is given by,

[tex]\rm v^2 = u^2 + 2as[/tex]

Rock X is released first followed by Rock Y, then Rock X has a greater speed than Rock Y therefore the distance covered by X is longer. This is because at any time 't' Rock X have already attained some velocity, its velocity v is more hence the separation 's' increases. Conclusively, 's' increases at all times that Rock X falls since Rock X falls with a greater velocity than Rock Y.

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https://brainly.com/question/19979064?referrer=searchResults