$1.3.8.$ A mortar is fired at objects located on the mountainside. At what distance from the mortar will the mines fall if their initial velocity is $v$, the angle of inclination of the mountain is $\alpha$, and the angle of fire relative to the horizon is $\beta$?
The equations of motion of the projectile can be written as follows: $$ {x}={v}_{0}{t}\cos{\beta} $$ $$ {y}={v}_{0}{t}\sin{\beta}-\frac{{g}{t}^{2}}{2} $$ Substitute into the equations the coordinates of the target $x = L; \;y = L \tan \alpha$ $$ {L=v_{0}t\cos\beta} $$ $$ Ltg\alpha=v_{0}t\sin\beta-\frac{gt^{2}}{2} $$ Let us express time from the first equation of the last system of equations and substitute its value into the second equation $$ {t=\frac{L}{v_{0}\cos\beta}} $$ $$ {Ltg\alpha=v_{0}\frac{L}{v_{0}\cos\beta}sin\beta-\frac{g}{2}\frac{L^{2}}{v_{0}^{2}\cos^{2}\beta} } $$ Where $$ {v}_{0}=\sqrt{\frac{{gL}\cos\alpha}{2\cos\beta\sin(\beta-\alpha)}} $$ We express $L$, $$ L = \frac{ 2\cos\beta\sin(\beta-\alpha)\cdot v^2_0}{g\cos\alpha} $$ And we find the flight range along the wall: $$ l = \frac{L}{\cos \alpha} $$ $$ \fbox{$l = \frac{ 2v^2_0}{g} \frac{ \cos\beta\sin(\beta-\alpha)}{\cos^2\alpha}$} $$
$$L=\frac{2v^2}g\frac{\cos^2\beta}{\cos\alpha}(\text{tg}\beta-\text{tg}\alpha)$$