The possible orbits of a body about the Sun in terms of their total mechanical energy If we assume that the Sun is at rest in an inertial reference frame, the total mechanical energy ( $E$ ) of the Sun and the orbiting body is constant and equal to the sum of the kinetic energy ( $\mathcal K$ ) and the gravitational potential energy ( $\mathcal U$ ).
Is there a mathematical-physical explanation, for students of an high school, with the specific steps because:

for a high speed object with $$\mathcal K>|\mathcal U|, \text { and } E>0$$ the orbit is unbounded and its trajectory is open or hyperbolic. When $$\mathcal K=|\mathcal U|, E = 0$$ the orbit is still unbounded but the trajectory is parabolic. For $$\mathcal K<|\mathcal U|,
E >0 $$ and the orbit is termed bounded with an elliptic trajectory. For $\mathcal K=0$ there is no orbit.

 A: The only way I know to show analytically that the trajectories are ellipses, parabolas, or hyperbolas involves solving a differential equation. In a comment, you explained that your students have only had some precalculus, so I don’t think you can demonstrate this to them, although you could certainly tell them that “it can be shown”.
If they understand that force determines acceleration, acceleration determines the change in velocity, and velocity determines the change in position, and if they know how to do some programming, they could write a computer program to numerically simulate trajectories. However, you might then get into problems with numerical error accumulation with the simplest algorithm. The trajectories may not be sufficiently accurate; for example, the elliptical orbits won’t close.
A very simple approach is just to talk about the energy equation,
$$E=\frac12mv^2-\frac{GMm}{r}=\text{const}$$
for a small mass $m$ moving in the field of a large mass $M$.
You can explain that when $E=0$, the small mass can just barely get to $r=\infty$ with zero velocity. And when $E<0$ it cannot get to $r=\infty$, because the kinetic term cannot be negative, so it must be in a bound orbit.
