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
Expert's answer
2020-06-04T09:59:01-0400

Let hh 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:


E1=W=mghE_1 = W = mgh

where m=0.1kgm = 0.1 kg is the mass of the apple and g=9.8m/s2g = 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:


E2=mv2/2+mg2E_2 = mv^2/2 + mg\cdot 2

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

According to the conservation energy law:


E1=E2E_1 = E_2mgh=mv2/2+mg2h=v2/2+2g=102/2+29.8=69.6mmgh = 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.


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