4. (20 points) A ball was thrown from the surface of the Earth at an angle of $45^{\circ}$ with a speed of $v_{0}=20 \mathrm{M} / \mathrm{s}$. How long will it take for the velocity vector of the ball to turn by an angle of $90^{\circ}$? Neglect air resistance. The acceleration due to gravity is $g=10 \mathrm{M} / \mathrm{s}^{2}$.
Answer: $\approx 2.83 s$.

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:

$$
y=v_{0 y} t-\frac{g t^{2}}{2}=v_{0} \sin 45^{\circ} t-\frac{g t^{2}}{2}=10 \sqrt{2} \cdot t-5 t^{2}=0
$$

From this, we obtain that $t=2 \sqrt{2} \approx 2.83 s$.