Answer to Question #120089 in Mechanics | Relativity for Amanda

Question

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?

Answer 1

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:

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

Answer. h = 69.6 m.