An object held in the air as a GPE of 470J....What is the object's kinetic energy?

Question

An object held in the air as a GPE of 470J. The object then is dropped. Half way down, what is the object's kinetic energy?

Answer 1

If GPE=470 Joules (J or j)
then
at half way down
KE=(1/2)GPE=.5(470)=235J
Why?
The total energy of the system is
KE+PE= (.5)mv^2 + mgh = GPE or PE(max)
KE - kinetic energy
PE - potential energy
m - mass of the object
v - velocity (or speed) of the object
g - local gravitational acceleration constant (in our case 9.81m/s^2)
and finally
h - the height above ground

since it went half way the total energy got split half way between kinetic and potential.

so 1/3 of th eway down KE=(1/3)GPE and PE =(2/3)GPE