Угол броска мяча

To solve this problem, we'll analyze the projectile motion of the ball considering both vertical and horizontal accelerations.

Given:

• Initial speed $v_0 = 10 \, \text{m/s}$
• Wind acceleration $a_{\text{wind}} = 15 \, \text{m/s}^2$ (opposite to the initial horizontal velocity)
• Gravitational acceleration $g = 10 \, \text{m/s}^2$

1. Determine the Time of Flight $T$:

The vertical motion is unaffected by the horizontal acceleration. The time to reach the maximum height is: $t_{\text{up}} = \frac{v_0 \sin\alpha}{g}$

The total time of flight is: $T = 2 t_{\text{up}} = \frac{2 v_0 \sin\alpha}{g}$

2. Set Up the Horizontal Motion Equation:

For the ball to return to the starting point, the horizontal displacement must be zero: $x(T) = v_0 \cos\alpha \cdot T + \frac{1}{2} (-a_{\text{wind}}) T^2 = 0$

Simplify: $v_0 \cos\alpha \cdot T = \frac{1}{2} a_{\text{wind}} T^2$

Divide both sides by $T$ (since $T > 0$): $v_0 \cos\alpha = \frac{1}{2} a_{\text{wind}} T$

3. Substitute $T$ from Step 1: $v_0 \cos\alpha = \frac{1}{2} a_{\text{wind}} \left( \frac{2 v_0 \sin\alpha}{g} \right)$

Simplify: $v_0 \cos\alpha = \frac{a_{\text{wind}} v_0 \sin\alpha}{g}$

Cancel $v_0$: $\cos\alpha = \frac{a_{\text{wind}} \sin\alpha}{g}$

4. Solve for $\tan\alpha$: $\cos\alpha = k \sin\alpha$ Where $k = \frac{a_{\text{wind}}}{g} = \frac{15}{10} = 1.5$.

Divide both sides by $\cos\alpha$: $1 = k \tan\alpha$ $\tan\alpha = \frac{1}{k} = \frac{1}{1.5} = \frac{2}{3}$

5. Calculate $\alpha$: $\alpha = \arctan\left(\frac{2}{3}\right) \approx 33.69^\circ$

6. Round to the Nearest Whole Number: $\alpha \approx 34^\circ$

Answer:

34