Question

two coconuts fall freely from rest at the same time, one twice as high as the other. If The coconut from the shorter tree takes 2.0 s to reach the ground, how long will it take the other coconut to reach the ground?

Answer 1

Take the first coconut's starting position to be the origin, and the downward direction to be positive. The first coconut's position is determined by

$y_1=\dfrac12gt^2$

where $g$ is the acceleration due to gravity.

So if it takes 2.0 s to reach the ground, then

$y_1=\dfrac12\left(9.8\,\dfrac{\mathrm m}{\mathrm s^2}\right)(2.0\,\mathrm s)^2=20.\,\mathrm m$

(rounding to 2 significant digits)

The second coconut starts 20 m higher than the first, so its initial displacement is -20 m relative to the origin, and its overall position over time is given by

$y_2=-20.\,\mathrm m+\dfrac12gt^2$

Reaching the ground is a matter of obtaining $y_2=20\,\mathrm m$, which requires a time of

$20\,\mathrm m=-20\,\mathrm m+\dfrac12\left(9.8\,\dfrac{\mathrm m}{\mathrm s^2}\right)t^2\implies t=2.9\,\mathrm s$