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=\frac12gt^2`

where
`g` is the acceleration due to gravity.

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

`y_1=\frac12\left(9.8\,(\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+\frac12gt^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+\frac12\left(9.8\,(\mathrm m)/(\mathrm s^2)\right)t^2\implies t=2.9\,\mathrm s`