From the story. Hickam's account says the boys learned to estimate altitude with physics as well as with their theodolites. Energy gives a second way to the same answer, which is a good check. (More in October Sky.)
At burnout the rocket has kinetic energy $\tfrac12 m v_b^2$. As it coasts up, that becomes gravitational potential energy. Without drag the two match exactly; with drag, some energy warms the air, and the apogee is lower.
Key idea
No drag: $\tfrac12 m v_b^2 = m g\,\Delta h$, so $\Delta h = \dfrac{v_b^2}{2g}$, the same as the kinematic answer. With drag: energy lost $= \tfrac12 m v_b^2 - m g\,(H - h_b)$, where $H$ is the tracked apogee.
Worked example: Where did the energy go?
A $200\ \text{g}$ rocket burns out at $40\ \text{m}$ moving $60\ \text{m/s}$, and is tracked at $180\ \text{m}$. How much energy did drag take?
- Kinetic at burnout: $\tfrac12(0.2)(60)^2 = 360\ \text{J}$.
- Potential gained: $0.2(9.8)(140) = 274\ \text{J}$.
- Drag took about $86\ \text{J}$, nearly a quarter.
Answer: $\approx 86\ \text{J}$
Common mistake
Measuring the potential energy from the ground instead of from burnout. Kinetic energy at burnout only pays for the climb after burnout.