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Sagot :
To solve the problem of determining the horizontal range of an apple thrown with a speed of [tex]\(30 \, \text{m/s}\)[/tex] at an angle of [tex]\(30^\circ\)[/tex] above the horizon, we follow these steps:
### Step 1: Break the initial velocity into horizontal and vertical components
Given:
- Initial speed, [tex]\(v_0 = 30 \, \text{m/s}\)[/tex]
- Angle of projection, [tex]\(\theta = 30^\circ\)[/tex]
- Acceleration due to gravity, [tex]\(g = 9.8 \, \text{m/s}^2\)[/tex]
First, we find the horizontal and vertical components of the initial speed:
[tex]\[ v_{\text{horizontal}} = v_0 \cos(\theta) \][/tex]
[tex]\[ v_{\text{vertical}} = v_0 \sin(\theta) \][/tex]
### Step 2: Calculate the horizontal component of the velocity
[tex]\[ v_{\text{horizontal}} = 30 \cos(30^\circ) \][/tex]
Cosine of [tex]\(30^\circ\)[/tex] is [tex]\(\frac{\sqrt{3}}{2}\)[/tex], so:
[tex]\[ v_{\text{horizontal}} = 30 \cdot \frac{\sqrt{3}}{2} \][/tex]
[tex]\[ v_{\text{horizontal}} \approx 30 \cdot 0.866 = 25.98 \, \text{m/s} \][/tex]
### Step 3: Calculate the vertical component of the velocity
[tex]\[ v_{\text{vertical}} = 30 \sin(30^\circ) \][/tex]
Sine of [tex]\(30^\circ\)[/tex] is [tex]\(\frac{1}{2}\)[/tex], so:
[tex]\[ v_{\text{vertical}} = 30 \cdot \frac{1}{2} \][/tex]
[tex]\[ v_{\text{vertical}} = 15 \, \text{m/s} \][/tex]
### Step 4: Determine the time of flight
The time to reach the highest point is calculated using:
[tex]\[ t_{\text{up}} = \frac{v_{\text{vertical}}}{g} = \frac{15}{9.8} \][/tex]
[tex]\[ t_{\text{up}} \approx 1.53 \, \text{seconds} \][/tex]
Since the time to reach the highest point and the time to come back down are equal, the total time of flight is:
[tex]\[ t_{\text{total}} = 2 \cdot t_{\text{up}} \][/tex]
[tex]\[ t_{\text{total}} \approx 2 \cdot 1.53 = 3.06 \, \text{seconds} \][/tex]
### Step 5: Calculate the horizontal range
The horizontal range is given by:
[tex]\[ R = v_{\text{horizontal}} \times t_{\text{total}} \][/tex]
[tex]\[ R = 25.98 \, \text{m/s} \times 3.06 \, \text{s} \][/tex]
[tex]\[ R \approx 79.53 \, \text{meters} \][/tex]
Thus, the horizontal range of the apple is approximately [tex]\(79.53 \, \text{meters}\)[/tex].
### Conclusion
Thus, the closest answer to the calculated horizontal range is:
[tex]\[ \boxed{80 \, \text{meters}} \][/tex]
Therefore, the correct option is:
[tex]\[ D. \, 80 \, \text{m} \][/tex]
### Step 1: Break the initial velocity into horizontal and vertical components
Given:
- Initial speed, [tex]\(v_0 = 30 \, \text{m/s}\)[/tex]
- Angle of projection, [tex]\(\theta = 30^\circ\)[/tex]
- Acceleration due to gravity, [tex]\(g = 9.8 \, \text{m/s}^2\)[/tex]
First, we find the horizontal and vertical components of the initial speed:
[tex]\[ v_{\text{horizontal}} = v_0 \cos(\theta) \][/tex]
[tex]\[ v_{\text{vertical}} = v_0 \sin(\theta) \][/tex]
### Step 2: Calculate the horizontal component of the velocity
[tex]\[ v_{\text{horizontal}} = 30 \cos(30^\circ) \][/tex]
Cosine of [tex]\(30^\circ\)[/tex] is [tex]\(\frac{\sqrt{3}}{2}\)[/tex], so:
[tex]\[ v_{\text{horizontal}} = 30 \cdot \frac{\sqrt{3}}{2} \][/tex]
[tex]\[ v_{\text{horizontal}} \approx 30 \cdot 0.866 = 25.98 \, \text{m/s} \][/tex]
### Step 3: Calculate the vertical component of the velocity
[tex]\[ v_{\text{vertical}} = 30 \sin(30^\circ) \][/tex]
Sine of [tex]\(30^\circ\)[/tex] is [tex]\(\frac{1}{2}\)[/tex], so:
[tex]\[ v_{\text{vertical}} = 30 \cdot \frac{1}{2} \][/tex]
[tex]\[ v_{\text{vertical}} = 15 \, \text{m/s} \][/tex]
### Step 4: Determine the time of flight
The time to reach the highest point is calculated using:
[tex]\[ t_{\text{up}} = \frac{v_{\text{vertical}}}{g} = \frac{15}{9.8} \][/tex]
[tex]\[ t_{\text{up}} \approx 1.53 \, \text{seconds} \][/tex]
Since the time to reach the highest point and the time to come back down are equal, the total time of flight is:
[tex]\[ t_{\text{total}} = 2 \cdot t_{\text{up}} \][/tex]
[tex]\[ t_{\text{total}} \approx 2 \cdot 1.53 = 3.06 \, \text{seconds} \][/tex]
### Step 5: Calculate the horizontal range
The horizontal range is given by:
[tex]\[ R = v_{\text{horizontal}} \times t_{\text{total}} \][/tex]
[tex]\[ R = 25.98 \, \text{m/s} \times 3.06 \, \text{s} \][/tex]
[tex]\[ R \approx 79.53 \, \text{meters} \][/tex]
Thus, the horizontal range of the apple is approximately [tex]\(79.53 \, \text{meters}\)[/tex].
### Conclusion
Thus, the closest answer to the calculated horizontal range is:
[tex]\[ \boxed{80 \, \text{meters}} \][/tex]
Therefore, the correct option is:
[tex]\[ D. \, 80 \, \text{m} \][/tex]
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