From the story. At Cape Coalwood the motor's roar lasted only moments, yet the rockets kept climbing long after it stopped. That climb with no push is the coast. (More in October Sky.)
Once the motor burns out, the only force (ignoring air) is gravity. The rocket keeps rising because of the speed it already has, losing $9.8\ \text{m/s}$ each second, until it stops for an instant at apogee, then falls.
Key idea
From burnout speed $v_b$ at height $h_b$: $$\role[coasting time]{time}{t_\text{coast}} = \frac{\role[burnout speed]{velocity}{v_b}}{\role[acceleration of gravity]{acceleration}{g}}, \qquad \role[apogee height]{position}{h_\text{apogee}} = \role[burnout height]{position}{h_b} + \frac{\role[burnout speed]{velocity}{v_b}^2}{2\role[acceleration of gravity]{acceleration}{g}}.$$ Falling from rest at apogee $H$: $t = \sqrt{2H/g}$. A model motor's delay (burnout to parachute) is chosen to be close to the coast time.
Worked example: Apogee
Burnout at $40\ \text{m}$ moving up at $50\ \text{m/s}$. How high, and when?
- Coast time $50/9.8 \approx 5.1\ \text{s}$.
- Rise $50^2/19.6 \approx 128\ \text{m}$, so apogee $\approx 168\ \text{m}$.
Answer: $\approx 168\ \text{m}$, $5.1\ \text{s}$ after burnout
Common mistake
Thinking the rocket starts to fall the moment the motor stops. It is still going up, just slowing.