Weg über die quadratische Ergänzung:
\( \frac{1}{2} \) * g * \( t^{2} \) - v * t + y = 0|-y
\( \frac{1}{2} \) * g * \( t^{2} \) - v * t = - y|*2
g * \( t^{2} \) - 2 v * t = - 2 y|:g
\( t^{2} \) - \( \frac{2v}{g} \) * t = - \( \frac{2}{g} \)* y
(t-\( \frac{v}{g} \))^2= - \( \frac{2}{g} \)* y+(\( \frac{v}{g} \))^2= - \( \frac{2g}{g^2} \)* y + \( \frac{v^2}{g^2} \)
(t-\( \frac{v}{g} \))^2=\( \frac{v^2-2g*y}{g^2} \)|\( \sqrt{} \)
1.)t - \( \frac{v}{g} \)= \( \frac{1}{g} \)*\( \sqrt{v^2-2g*y} \)
t₁ = \( \frac{v}{g} \) + \( \frac{1}{g} \)*\( \sqrt{v^2-2g*y} \)
2.)t - \( \frac{v}{g} \)= -\( \frac{1}{g} \)*\( \sqrt{v^2-2g*y} \)
t₂ = \( \frac{v}{g} \) - \( \frac{1}{g} \)*\( \sqrt{v^2-2g*y} \)