Projectile Motion

The general projectile motion of an oblique throw of a body in the Earth's gravitational field and includes horizontal and vertical throws as special cases. In an oblique throw, the body is thrown at an angle $\alpha$ to the Earth's surface from a certain height $h$ with an initial velocity $v_0$.

First, the velocity vector $\vec{v}_0$ can be divided into its $x$ and $y$ components using trigonometric functions. i.e., the following substitutions are made:
\begin{eqnarray}
v_{0,x} = v_0\cos\varphi\\
v_{0,y} = v_0\sin\varphi
\end{eqnarray}

Furthermore, the same assumptions apply as for a horizontal throw: the velocity component $v_{0,x}$ in the $x$-direction remains constant, while the component $v_{0,y}$ in the $y$-direction increases at every instant, since the body is accelerated in this direction.

The equations of motion are then as follows:
\begin{eqnarray}
x(t) &=& v_0\cos\varphi\,t\\
y(t) &=& -\frac{1}{2}gt^2 + v_0\sin\varphi\,t + h
\end{eqnarray}
If we eliminate time in the second equation by rearranging the first equation, we get the following quadratic relationship:
\begin{equation}
{y(x) = -\frac{1}{2}g\frac{x^2}{v_0^2\cos^2\varphi} + x\tan\varphi +h}
\end{equation}
Here, the relationship $\tan\varphi = \sin\varphi/\cos\varphi$ was used.
Due to the quadratic relationship between throwing distance $x$ and throwing height $y$, the resulting curve is called a throwing parabola.

The maximum throwing distance and height can, in principle, be derived from this derived relationship.

Due to the complexity of the individual terms, however, it is recommended to use the system of equations established at the beginning. mTo determine the maximum throwing distance, $y(t)$ is again set equal to 0:

\begin{equation}
y(t) = 0 \Leftrightarrow -\frac{1}{2}gt^2 + v_0\sin\varphi\,t + h = 0
\end{equation}
This equation can now be solved, for example, using the $pq$ formula. The result is the time $t_W$ that elapses until the maximum throwing distance is reached:

\begin{equation}
t_\mathrm{W} = \frac{\sqrt{\sin^2\varphi\,v_0^2 + 2h}+v_0\sin\varphi}{g}
\end{equation}

The throwing distance is then determined by substituting $x(t)$ into $x(t)$.

The maximum height can also be calculated analogously. This is achieved when the velocity in the $y$-direction reaches 0.
Mathematically speaking, the derivative of the function $y(t)$ with respect to time must vanish, i.e., the following must hold:
\begin{equation}
y'(t) = 0 \Leftrightarrow -gt + v_0\sin\varphi = 0
\end{equation}
Since this is a linear equation, the time $t_\mathrm{H}$ after which the maximum height is reached is:
\begin{equation}
{t_\mathrm{H} = \frac{v_0}{g}\sin\varphi}
\end{equation}
Substituting the time into the function $x(t)$ gives the distance after which the maximum throwing height is reached:
\begin{equation}
{x_H = \frac{v_0^2}{g}\sin\varphi\,\cos\varphi}
\end{equation}
Similarly, time can also be inserted into the function $y(t)$ to calculate the maximum height:
\begin{equation}
{y_H = \frac{v_0^2\sin^2\varphi}{2g} + h}
\end{equation}