1. **State the problem:** An object of mass 5 kg is dropped from a height of 100 m. We need to find the height at which its kinetic energy (KE) is twice its potential energy (PE). 2. **Formulas and rules:** - Potential energy at height $h$ is given by $PE = mg h$ where $m$ is mass, $g$ is acceleration due to gravity, and $h$ is height. - Kinetic energy is $KE = \frac{1}{2} m v^2$. - Total mechanical energy at the start (height 100 m) is $E = mg \times 100$. - Since energy is conserved, $E = PE + KE$ at any height. 3. **Set up the equation:** We want $KE = 2 PE$. Using conservation of energy: $$mg \times 100 = PE + KE$$ Substitute $KE = 2 PE$: $$mg \times 100 = PE + 2 PE = 3 PE$$ So, $$PE = \frac{mg \times 100}{3}$$ 4. **Calculate height $h$:** Since $PE = mg h$, $$mg h = \frac{mg \times 100}{3}$$ Cancel $mg$ on both sides: $$h = \frac{100}{3} \approx 33.33$$ 5. **Answer:** The height at which the kinetic energy is twice the potential energy is approximately 33.33 meters.