From the story. Air drags on every rocket, including the ones that flew from Cape Coalwood, so real rockets never quite reach the heights the simple formulas promise. (More in October Sky.)
Moving through air costs force. For a rocket the drag is close to $F_D = \tfrac12\rho C_D A v^2$: double the speed and drag grows about four times. It always points against the motion: down on the way up, up on the way down.
With drag the acceleration depends on the speed, so there is no simple formula for apogee. Instead, take small time steps. This is Euler's method: use the present speed to find the acceleration, then move forward a little.
Key idea
Coasting up: $a = -g - \dfrac{k v^2}{m}$ with $k = \tfrac12\rho C_D A$. One Euler step of size $\Delta t$: $$h_{n+1} = h_n + v_n\Delta t, \qquad v_{n+1} = v_n + a_n\Delta t.$$ Smaller steps give better answers.
Worked example: One step
A $250\ \text{g}$ rocket coasts up at $70\ \text{m/s}$ with $k = 0.0012\ \text{kg/m}$. Take one step of $0.1\ \text{s}$.
- $a = -9.8 - 0.0012(70)^2/0.25 = -9.8 - 23.5 = -33.3\ \text{m/s}^2$.
- $v_1 = 70 - 3.33 = 66.7\ \text{m/s}$; $h$ grows by $70(0.1) = 7\ \text{m}$.
Answer: $v_1 \approx 66.7\ \text{m/s}$
Common mistake
Adding drag in the wrong direction, or using the starting speed's drag for the whole flight.