$1.3.5.$ A stone is thrown at a velocity $v$ at an angle $\varphi$ to the horizon. After what time will the velocity be at the angle $\alpha$ with the horizon?
The horizontal component of velocity remains unchanged: $$ v_{x}(\varphi) = v_{x}(\alpha) = v \cdot \cos{\varphi} $$ And the horizontal component decreases, depending on time, according to the law: $$ v_{y}(t) = vt \sin{\varphi} - gt $$ From where the angle that the velocity makes with the horizon is determined as: $$ \tan{\alpha} = \frac{v_{y}(t)}{v_x} $$ Or, $$ \tan{\alpha} \cdot v \cdot \cos{\varphi} = v \cdot \sin{\varphi} - gt $$ From where we obtain the required moment of time: $$ \fbox{$t= \frac{v}{g}(\sin{\varphi} - \cos{\varphi}\tan{\alpha})$} $$
$$t= \frac{v}{g}(\sin{\varphi} - \cos{\varphi}\tan{\alpha})$$