free fall - freier Fall
$$y(t) = - \frac{1}{2}gt^2$$
${g = 9.81 \frac{m}{s^2}}$ and y(t) is the vertical position in time starting at the origin (0,0)
free fall with horizontal motion - waagrechter Wurf
$$x(t) = v_0 \cdot t $$
$$y(t) = - \frac{1}{2}gt^2$$
projectile motion - schräger Wurf
$$x(t) = v_0 \cdot \cos{\alpha} \cdot t$$
$$y(t) = v_0 \cdot \sin{\alpha} \cdot t - \frac{1}{2}gt^2$$
${g = 9.81 \frac{m}{s^2}}$, $\alpha$ is the inclination angle and ${v_0}$ is the magnitude of the velocity.
The horizontal component is ${v_0 \cos{\alpha}}$.
The vertical component is ${v_0 \sin{\alpha}}$.
example - given distance of the entire motion and the angle
Here, the approach is to set y = 0 and solve for the time first.
Then, substitute the time in terms of $v_0$ into the equation with the horizontal distance to solve for $v_0$ .
$$0 = v_0 \cdot \sin{\alpha} \cdot t - \frac{1}{2}gt^2$$
$${\Rightarrow 0 = t \cdot \left(v_0 \cdot \sin{\alpha} - \frac{1}{2}gt\right)}$$
$${\Rightarrow t = 0 \wedge v_0 \cdot \sin{\alpha} - \frac{1}{2}gt = 0}$$
$$\Rightarrow t = 0 \wedge v_0 \cdot \sin{\alpha} = \frac{1}{2}gt$$
$$\Rightarrow t = 0 \wedge t = \frac{2 \cdot v_0 \cdot \sin{\alpha}}{g}$$
Substituting $t = \frac{2 \cdot v_0 \cdot \sin{\alpha}}{g}$ into $x(t) = v_0 \cdot t \cdot \cos{\alpha}$ gives us
$$x(t) = v_0 \cdot \frac{2 \cdot v_0 \cdot \sin{\alpha}}{g} \cdot \cos{\alpha}$$
$$\frac{x(t) \cdot g}{ 2 \cdot \sin{\alpha}\cdot \cos{\alpha}} = v_0^2 $$
Hence,
$$ v_0 = \sqrt{\frac{x(t) \cdot g}{ 2 \cdot \sin{\alpha}\cdot \cos{\alpha}}} $$