Horizontal Throw

In a horizontal throw, the body is thrown parallel to the ground. The vertical and horizontal motion of a body in the Earth's gravitational field can be considered independently of each other.

The horizontal throw of a body is characterized by the fact that the body has a time-independent velocity $v_{0,x}$ in the horizontal $x$-direction. In the vertical $y$-direction, it experiences a constant acceleration $g$ perpendicular to the ground, i.e., the velocity $v_y(t)$ starts at 0 m/s and increases continuously over time with $v_y(t) = gt$.
The equations of motion of the body dropped at a height $h$ in the positive $x$ direction can be written in the following form using the general path-time law from the general equation of motion:

\begin{eqnarray}
x(t) &=& v_{0,x} t\\
y(t) &=& -\frac{1}{2}gt^2 + h
\end{eqnarray}

To simplify the representation, the upper equation can be rearranged for $t$ and then substituted into the lower equation.
The resulting function then describes the right branch of a parabola:

\begin{equation}
{y(x) = h-\frac{1}{2}g\frac{x^2}{v_{0,x}^2}}
\end{equation}

The magnitude of the orbital velocity $v(t)$ is given at any point in time $t$ using the Pythagorean theorem as follows:

\begin{equation}
|\vec{v}(t)| = \sqrt{v_{0,x}^2 + g^2t^2}
\end{equation}

Furthermore, the angle between the velocity vector and the horizontal can be calculated using the following formula:

\begin{equation}
\tan\varphi = \frac{v_y(t)}{v_x(t)}
\end{equation}

After substituting $v_y$ and $v_y$, this results in:

\begin{equation}
{\tan\varphi = \frac{gt}{v_{0,x}}}
\end{equation}

Using the above equations, a horizontal throw in the Earth's gravitational field can be fully described, neglecting air resistance.