You are exploring a planet and drop a small rock from the edge of a cliff. In coordinates where the +y direction is downward and

Answer:

The gravitational acceleration on the planet is 5.00 m/s²

Explanation:

This situation resembles a free-fall scenario, but with a different gravitational constant; the provided equation is as follows:

       y-yo = ½ gₐ t²       (1)

They mention creating a squared time graph illustrating the distance variation, which is beneficial for linearizing a nonlinear curve - specifically, by plotting the axis raised to a power, in this case, the square against another variable.

    Let's continue with the analysis, noting we have a linear relationship; we can write the line’s equation:

        y1 = m x1 + b       (2)

Here, “y1” is the dependent variable, “x1” is the independent variable, “m” is the slope, and “b” indicates the intercept

Since the stone is released with zero initial velocity, we observe that b = 0,

On the y-axis, we label time squared “t²”, and on the x-axis, we mark “y-yo.” For clarity, we can rewrite equation 1 as follows:

        t² = 2 /gₐ  (y-yo)

Aligning the two expressions allows us to relate the terms correspondingly

        y1 = t²

        x1 = (y-yo)

        m = 2/gₐ

        b= 0

After substituting, we find

        m = 2/gₔ[/color=red.entries():count].

        gₐ = 2/m

        gₐ = 2/ 0.400

        gₐ = 5.00 m/s²

Thus, the acceleration due to gravity measures at 5.00 m/s² on the planet, while being conscientious of significant figures