A stone is thrown from the surface of the earth vertically upward at a speed of 12 m / s. Determine at what height

A stone is thrown from the surface of the earth vertically upward at a speed of 12 m / s. Determine at what height its kinetic energy will decrease by 2 times. neglect the resistance to movement. take the coefficient g equal to 10n / kg

The expression for determining the kinetic energy is:
E = m * v² / 2
At the beginning of our movement, the kinetic energy is equal to:
E = m * v0² / 2
The moment the body moves:
E1 = m * v² / 2
Let’s divide E by E1:
E / E1 = (m * v0² / 2) / (m * v² / 2) = v0² / v²
v0 / v = (E / E1) ² = 4
v = v0 / 4
The equation of motion of a body thrown vertically upwards from the earth’s surface is as follows:
h = v0 * t-g * t² / 2
Speed ​​equation:
v = v0-g * t / 2
Let us express time from this expression:
t = 2 * (v0-v) / g = 2 * (4 * v0 / 4-v0 / 4) / g = 1.5 * v0 / g
Substitute the time in the expression for h:
h = v0 * t-g * t² / 2 = v0 * (1.5 * v0 / g) -g * (1.5 * v0 / g) ² / 2
Substitute the numbers, find the height:
h = v0 * (1.5 * v0 / g) -g * (1.5 * v0 / g) ² / 2 = 12 * (1.5 * 12/10) -10 * (1.5 * 12 / 10) ² / 2 = 21.6-16.2 = 5.4 m.
Answer: at a height of 5.4 m, the kinetic energy will decrease by 2 times.