I always have been excited about fireworks, beacuse they are a mixture of art and phisics. Beying myself a musician and also not a phisician but close, I think there is a wonderful marriage between both.
Phisics are not the hardest part - artistic is. If you ever learned basic phisics, like Newton's gravity laws, Potential Energy and Energy Conservation you are half way to do everything.
Some notes on that.
The speed, position and acceleration of a body can be calculated easily. You have few forces applied to a firework star. Those are:
Let me give you an example:
In the first stage (0) you want to put your star up high in the sky so that when stage 1 comes the star is standing still. To stand still means that its velocity equals zero.
You can choose your forward thrust, the duration of the stage and the initial velocity.
Lets call t the time in seconds that elapses between the firing and the point star stands still.
Lets call Vo the initial velocity.
Lets call g the gravity force (that equals about 34ft/s²)
Lets call F the initial thrust.
You have to choose t, Vo and F. Lets take a look at the general equations for velocity first.
The equation for a particle's velocity can be calculated by V = Vo + a.t for each axis, where a is the sum of all vectorial forces when projected on that same axis. As in the 1st stage we only have one axis (upward/downward) it becomes very simple. Thurst is positive (upward) and gravity is negative (downward). So a equals F -g
This makes V = Vo + (F-g).t
You want your speed to be zero, so
0 = Vo + (F-g).t
You have three undetermined variables so you have to pick a value for two of them. Lets choose t first. Lets define that we want the star to be positioned after 4 seconds. That makes:
0 = Vo + (F-g).4 , which equals to say 0 = Vo + 4.F - 4.g .
4.g is a known value, is 4 times 34, that makes 136. So,
0 = Vo + 4.F - 136 , that makes -Vo = 4.F - 136 => Vo = 136 - 4.F
So if you dont want to use thrust (F=0), make Vo=136.
Remember Vo cannot be negative.
So there is not much you can change here, but.... how high in the sky will it be?
Acceleration is the difference between velocity, as velocity is the difference between positions. Latter is easy to explain: if you run for 100 feet in 1 second, your speed is 100ft/s (hundred feet per second). If you do it in 2 seconds, your speed will be 50ft/s.
So let's integrate the general velocity equation (integrate is the inverse of differentiate, so if you have an equation that speaks about difference (like velocity) you integrate).
That makes:
(integral sign here) V = Vo + a.t => X = Xo + Vo.t + ½ a.t²
Xo is the initial position. Lets say it starts at 0 feet :)
So your final position will be in the example:
X = 0 + 136*t + ½ (0-g) . t² (0-g) that is a=F-g and your thrust F was zero too.
So,
X = 136*4 => X = 544 feet .