$1.3.11.$ From the opening of the hose covered with a finger, two jets are shot at an angle $\alpha$ and $\beta$ to the horizon with the same initial velocity $v$. At what horizontal distance from the hole will the jets intersect?
$$ vt_{1} \cdot \cos \alpha = vt_{2} \cdot \cos \beta $$ $$ vt_{1}\cdot \sin\alpha - \frac{gt_{1}^{2}}{2}=vt_{2}\cdot \sin\beta-\frac{gt_{2}^{2}}{2} $$
From the first equation,
$$ t_{1}=t_{2} \cdot \frac{\cos \beta}{\cos \alpha} $$We substitute $t_{1}$ into the second equation and express $t_{2}$. Trigonometric formulas 63 will come in handy. The $t_{2}$ we have already obtained is enough to insert into $x=vt_{2}cos\beta$, where $x$ is the desired distance.
Using trigonometric formulas:
$$ t_{2}= \frac{2\nu ^{2}}{g(tg\beta +tg\alpha )cos\beta } $$ $$ \fbox{$x = \frac{2\nu ^{2}}{g(tg\beta +tg\alpha )}$} $$$$x = \frac{2\nu ^{2}}{g(tg\beta +tg\alpha )}$$