Угол броска мяча

Question

Мяч брошен с земли со скоростью v0 = 10 м/с под углом a к горизонту. На мяч во время полета действует встречный горизонтальный ветер, сообщая мячу постоянное ускорение a = 15 м/с2 в горизонтальном направлении. Известно, что мяч спустя некоторое время, после начала движения, вернулся в исходную точку. Пренебрегая силой сопротивления воздуха и считая, что ускорение свободного падения равно g = 10 м/с2, определите угол α. Ответ дайте в градусах, округлив его до целых.

Réflexion

Setting up the problem

I’m piecing together a projectile motion problem where a ball is thrown with a given initial speed and angle. A constant acceleration due to a headwind also affects the horizontal motion. The goal is to determine the initial angle α.

Piecing together the scenario

I’m working through the trajectory of a ball launched horizontally with a wind blowing against it, aiming to find the launch angle needed for it to return to the start. Gravity and wind are also factors.

Analyzing projectile motion

I’m piecing together equations for a projectile's horizontal and vertical motion, considering initial velocities and accelerations due to wind and gravity. The focus is on determining the time of flight T when the projectile returns to y = 0.

Assessing horizontal motion

OK, let me see. I’m focusing on how horizontal acceleration affects horizontal motion, ensuring x(T) hits 0. Moving forward, I’ll pinpoint the time of flight T.

Calculating flight time

I’m mapping out the time of flight for a projectile by analyzing the vertical and horizontal components, ultimately determining that the total time is $T = \frac{2v_0 \sin \alpha}{g}$ .

Determining α

By equating and substituting expressions for time and velocity, I simplified to $\cos \alpha = 1.5 \sin \alpha$ , leading to $\tan \alpha = 2/3$ .

Réponse

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: $\alpha \approx 34^\circ$

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