4. (20 points) A ball was thrown from the surface of the Earth at an angle of $30^{\circ}$ with a speed of $v_{0}=20 \mathrm{M} / \mathrm{c}$. How long will it take for the velocity vector of the ball to turn by an angle of $60^{\circ}$? Neglect air resistance. The acceleration due to gravity is $g=10 \mathrm{M} / \mathrm{c}^{2}$.
Answer: $t=2 s$

Solution. In fact, the problem requires finding the total flight time of the ball. The $y$-coordinate of the ball changes according to the law: $y=v_{0 y} t-\frac{g t^{2}}{2}=v_{0} \sin 30^{\circ} t-\frac{g t^{2}}{2}=10 \cdot t-5 t^{2}=0$

From this, we obtain that: $t=2 s$.