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 neglecting air resistance, the vertical displacement of an object released from rest is given by y − y0 = 1 2 gplanett2, where gplanet is the acceleration due to gravity on the planet. You measure t in seconds for several values of y − y0 in meters and plot your data with t2 on the vertical axis and y − y0 on the horizontal axis. Your data is fit closely by a straight line that has slope 0.400 s2/m. Based on your data, what is the value of gplanet?

They tell us that they make a squared time graph with the variation of the distance, it is appropriate to clarify this in a method to linearize a curve, which is plotted the nonlinear axis to the power that is raised, specifically, the linearization of a curve The square is plotted against the other variable.

Let's continue our analysis, as we have a linear equation, write the equation of the line.

y1 = m x1 + b (2)

where “y1” the dependent variable, “x1” the independent variable, “m” the slope and “b” the short point

In this case as the stone is released its initial velocity is zero which implies that b = 0,

We plot on the “y” axis the time squared “t²” and on the horizontal axis we place “y-yo”. To better see the relationship we rewrite equation 1 with this form

t² = 2 /gₐ (y-yo)

With the two expressions written in the same way, let's relate the terms one by one

y1 = t²

x1 = (y-yo)

m = 2/gap

b= 0

We substitute and calculate

m = 2/gp

gₐ = 2/m

gₐ = 2/ 0.400

gₐ = 5.00 m / s²

This is the value of the acceleration of gravity on the planet, note that the decimals are to keep the figures significant