You are exploring a planet and drop a small rock from the edge of a cliff. In co

Question

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 = 1/2gplanett2

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 0.700 s2/m. Based on your data, what is the value of gplanet?

Explanation / Answer

Assuming that the gplanet be constant on that planet near groung.

graph is between t2 and (y-y0) where t is time and (y-y0) is the displacement in vertical direction.

We know that S= ut + 1/2 at2 where S = displacement , u= initial velocity, a= acceleration

In this case u= 0, and slope is 0.700 s2/m i.e. slope = t2/(y-y0) which is equal to 2* gplanet here so

gplanet = 0.700/2=0.350 m/s2

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