# Answer to Question #120089 in Mechanics | Relativity for Amanda

Question #120089
A 0.10 kg apple falls from the top of a tree. After falling from the tree, the apple is measured to be traveling at 10 m/s when it is 2.0 m above the ground. How tall is the tree?
1
2020-06-04T09:59:01-0400

Let "h" be the height of the tree. Then the total energy of the apple at the top of the tree will be eaual to it's potential energy:

"E_1 = W = mgh"

where "m = 0.1 kg" is the mass of the apple and "g = 9.8m\/s^2" is the free-fall acceleration due to gravity.

At the point 2.0 m above the ground the energy of the apple will be the sum of the kinetic and potential energies:

"E_2 = mv^2\/2 + mg\\cdot 2"

where "v = 10m\/s" is the speed of the apple at that point.

According to the conservation energy law:

"E_1 = E_2""mgh = mv^2\/2 + mg\\cdot 2\\\\\nh = v^2\/2 + 2g = 10^2\/2 + 2\\cdot 9.8 = 69.6 m"

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