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

Answer:

C) True. The distance S increases over time, with v₁ = gt and v₂ = g (t-t₀), illustrating that v₁> v₂ for the same t.

Explanation:

We have a set of statements to evaluate for correctness. The most effective approach is to examine the problem in detail.

Using the equation for vertical launch, we acknowledge that the positive direction signifies downward movement.

Stone 1

    y₁ = v₀₁ t + ½ g t²

    y₁ = 0 + ½ g t²

Stone 2

Released shortly thereafter, let's assume a delay of one second, we can utilize the same timing mechanism

     t ’= (t-t₀)

    y₂ = v₀₂ t ’+ ½ g t’²

    y₂ = 0 + ½ g (t-t₀)²

    y₂ = + ½ g (t-t₀)²

We can now calculate the separation distance between the two stones, which is applicable for t> = to

    S = y₁ -y₂

    S = ½ g t²– ½ g (t-t₀)²

    S = ½ g [t² - (t² - 2 t t₀ + t₀²)]  

    S = ½ g (2 t t₀ - t₀²)

    S = ½ g t₀ (2 t - t₀)

This represents the distance between the two stones over time, with the coefficient outside the parentheses being constant.

For t < to, the first stone remains stationary while the distance grows.

For t > = to, the expression (2t/to-1) yields a value greater than 1, indicating that the distance expands as time progresses.

We can now analyze the different statements

A) false. The height difference increases over time.

B) False S increases.

C) It is true that S increases over time, with v₁ = gt and V₂ = g (t-t₀) indicating v₁> v₂ at the same t.