Free Fall
The quantities and formulas established so far will now be illustrated using the example of throws. All bodies in the Earth's gravitational field near the Earth's surface are subject to the same acceleration $a=-g$ in free fall, which is directed toward the Earth's center. The quantity $g$ (from the English word "gravity") is therefore also called the acceleration due to gravity. The minus sign in front of $g$ reflects the fact that the acceleration is directed downward from an observer standing on Earth.
It should be noted that the acceleration due to gravity varies at different locations on Earth. However, the deviations between the poles and the equator are only a few percent. In Central Europe, the value $g=9.81\,\mathrm{m}/\mathrm{s}^2$ or $g\approx 10\,\mathrm{m/s^2}$ can be used as a good approximation. For very large distances from the Earth's surface, Newton's law of gravity must be used to calculate acceleration instead of this approximation. The figure below includes all important parameters
Free fall occurs when an object falls freely downwards from a certain height $h$ and initially has no initial velocity $v_0$. The general distance-time law then gives the following for the distance traveled in the $y$ direction:
\begin{equation}
y(t) = -\frac{1}{2}gt^2 + h
\end{equation}
The fall time can be calculated directly from this equation. The ground is reached when $y(t) = 0$. This results in the equation
\begin{equation}
y(t) = 0\Leftrightarrow -\frac{1}{2}gt^2 + h = 0
\end{equation}
which, when rearranged for $t$, yields the fall time $t_\mathrm{F}$:
\begin{equation}
t_\mathrm{F} = \sqrt{\frac{2h}{g}}
\end{equation}
Since negative times are physically meaningless, only the positive solution of the quadratic equation is considered.
Vertical Throw
A vertical throw occurs when a body is thrown vertically upwards and has a defined initial velocity $v_0$ at the moment of release. All important parameters are defined in the following figure:
If it is dropped from a height of $h$, its motion in the $y$ direction can be described using the general distance-time law as follows:
\begin{equation}
{y(t) = -\frac{1}{2}gt^2 + v_0t + h}
\end{equation}
After being dropped, the body is decelerated by the acceleration due to gravity and reaches a maximum height before falling back down. It reaches the ground when the condition $y(t) = 0$ is met, i.e.
\begin{equation}
y(t) = 0\Leftrightarrow -\frac{1}{2}gt^2 + v_0t + h = 0
\end{equation}
This quadratic equation can be solved using the $pq$ formula. This gives the formula for the total flight time:
\begin{equation}
{t_\mathrm{F} = \frac{v_0}{g}+ \sqrt{\frac{v_0^2}{g^2} + \frac{2h}{g}}}
\end{equation}
The highest point at which the motion reverses can be determined in a similar way. To do this, the initial equation must be differentiated with respect to $t$ to obtain the speed as a function of time:
\begin{equation}
v(t) = gt + v_0
\end{equation}
The highest point is reached when $v(t) = 0$, so the flight time until reaching the highest point can be calculated by converting to $t$:
\begin{equation}
t_\mathrm{H} = \frac{v_0}{g}
\end{equation}
Substituting $t_\mathrm{H}$ back into $y(t)$, we obtain the maximum altitude:
\begin{equation}
y_\mathrm{H} = -\frac{1}{2}\frac{v_0^2}{g} + \frac{v_0^2}{g} + h
\end{equation}