Variant 134. From 10 to 20 minutes (including 10 and 20).

Grading criteria: 20 points - correct (not necessarily the same as above) solution and correct answer; $\mathbf{1 0}$ points - the solution is reduced to correct inequalities with respect to $t$, but further errors are made; 5 points - the equation and two inequalities with respect to poorly chosen unknown quantities are correctly written, the solution of which is either not obtained or obtained with errors.

## Problem 2 (Variant 131)

From a point on the surface of the earth, a large number of small balls are launched in all directions at the same speed of $10 \mathrm{m} / \mathrm{c}$. Among all the balls that land at a distance from the starting point not closer than $96 \%$ of the distance at which the farthest flying ball lands, find the one that will spend the most time in flight. What is this time? Express the answer in seconds and round to one decimal place. The acceleration due to gravity is $10 \mathrm{~m} / \mathrm{c}^{2}$.
Solution. The range of a body thrown with an initial velocity $V_{0}$ at an angle $\alpha$ to the horizontal is $l=\frac{V_{0}^{2} \sin 2 \alpha}{g}$, and the time of flight is $\tau=\frac{2 V_{0} \sin \alpha}{g}$.

According to the condition $\frac{V_{0}^{2} \sin 2 \alpha}{g} \geqslant \frac{96}{100} \frac{V_{0}^{2} \sin \left(2 \cdot 45^{\circ}\right)}{g}$, i.e., $\sin 2 \alpha \geqslant \frac{96}{100}$. Among these balls, we need to choose the one for which $\tau=\frac{2 V_{0} \sin \alpha}{g}$ is maximized, i.e., the one for which $\sin \alpha$ is the largest. Since $\arcsin \frac{24}{25} \leqslant 2 \alpha \leqslant \pi-\arcsin \frac{24}{25}$, then $\frac{1}{2} \arcsin \frac{24}{25} \leqslant \alpha \leqslant \frac{\pi}{2}-\frac{1}{2} \arcsin \frac{24}{25}$. Therefore, the maximum flight time is $\frac{2 V_{0}}{g} \sin \left(\frac{\pi}{2}-\frac{1}{2} \arcsin \frac{24}{25}\right)=\frac{2 V_{0}}{g} \cos \left(\frac{1}{2} \arcsin \frac{24}{25}\right)$

$=\frac{2 V_{0}}{g} \sqrt{\frac{1+\cos \left(\arcsin \frac{24}{25}\right)}{2}}=\frac{2 V_{0}}{g} \sqrt{\frac{1+\frac{7}{25}}{2}}=\frac{2 V_{0}}{g} \sqrt{\frac{16}{25}}=1.6 \text{s}$.
Answer: 1.6 seconds.

Answers of other options: