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?
Expert's answer

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"

Answer. h = 69.6 m.

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