From the story. Nobody can read a speedometer on a small rocket, at Cape Coalwood or anywhere else. What you can know is the motor's thrust and how long it burns, and from those come the speed and height at burnout. (More in October Sky.)
In the simplest model, the thrust is steady and the rocket's mass changes little, so the net acceleration $a = (T - mg)/m$ is constant during the burn. Then the familiar kinematic equations apply from liftoff to burnout. (Real motors vary their thrust and lose mass, so this is an estimate.)
Key idea
From rest, for a burn time $t_b$: $$\role[burnout speed]{velocity}{v_b} = \role{acceleration}{a}\,\role[burn time]{time}{t_b}, \qquad \role[burnout height]{position}{h_b} = \tfrac12 \role{acceleration}{a}\,\role[burn time]{time}{t_b}^2, \qquad a = \frac{T - mg}{m}.$$
Worked example: Burnout
A $90\ \text{g}$ rocket has a steady $5.0\ \text{N}$ thrust for $1.2\ \text{s}$. Find its burnout speed and height.
- $a = (5.0 - 0.882)/0.090 \approx 45.8\ \text{m/s}^2$.
- $v_b = 45.8(1.2) \approx 55\ \text{m/s}$; $h_b = \tfrac12(45.8)(1.2)^2 \approx 33\ \text{m}$.
Answer: $\approx 55\ \text{m/s}$, $\approx 33\ \text{m}$
Common mistake
Using $T/m$ as the acceleration. Subtract the weight first.