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Alexus [3.1K]
3 years ago
5

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 s

ame 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.
Physics
1 answer:
babunello [35]3 years ago
7 0

Answer:

Separation increases at all times that rock X falls because it falls with a greater speed

Explanation:

For both rocks, let initial velocity ∪=0

To find the displacement at any given time interval of Δt then

S= ∪Δt +0.5gΔt²

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

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