The path of a model rocket can be modelled using the following equation:
x = h-sin(theta) y=0.5*h-cos(theta) z=-10+sqrt((2^3)*g*h)/9.81, where (x, y, z) is in meters and time t is in seconds. The vertical component of velocity decreases exponentially with height while the horizontal component remains constant at zero until reaching maximum altitude.
The height of the model rocket when it reaches maximum altitude can be found using this equation:
h=sqrt((x^*)/g) h is in meters and x is in meters. After reaching its apogee, a model rocket will fall back to earth at a rate that’s proportional to time squared (in seconds). This means that twice the amount of time after liftoff will result in four times as much distance fell.
For example, if t equals 100 seconds, one-quarter of the way down from max altitude, we’ll have travelled about 150 m away from our launch point while travelling 150 m/s vertically downward. The total energy available for an object on Earth comes from gravity or electrical potential.