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

Explanation / Answer

the given equation is,

(y-yo)=1/2*(g_planet)*t^2

here,

acceleration due to planet is,

g_planet=2*(y-yo)/t^2

=1/(t^2/2*(y-yo))

=2/(t^2/(y-yo))

=2/slope

=2/[(0.9)sec^2/m]

===>

g_planet=2.22 m/sec^2

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