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Sagot :
To solve this problem, we'll use the principles of projectile motion. We'll break it down into steps to find the angle of projection, the time of flight, the range, and the maximum height.
### Step-by-Step Solution:
#### Given:
- Initial velocity (v) = 250 m/s
- Acceleration due to gravity (g) = 10 m/s²
- The maximum height (H) is one-third of the range (R)
#### i. Angle at which the bullet is fired (θ)
1. Equations of Projectile Motion:
- Maximum height (H):
[tex]\[ H = \frac{v^2 \sin^2(\theta)}{2g} \][/tex]
- Range (R):
[tex]\[ R = \frac{v^2 \sin(2\theta)}{g} \][/tex]
- Given relationship:
[tex]\[ H = \frac{R}{3} \][/tex]
2. Substituting H = R/3 into equation:
[tex]\[ \frac{v^2 \sin^2(\theta)}{2g} = \frac{1}{3} \left( \frac{v^2 \sin(2\theta)}{g} \right) \][/tex]
Simplifying:
[tex]\[ \frac{\sin^2(\theta)}{2} = \frac{\sin(2\theta)}{3} \][/tex]
3. Using the identity [tex]\(\sin(2\theta) = 2 \sin(\theta) \cos(\theta)\)[/tex]:
[tex]\[ \frac{\sin^2(\theta)}{2} = \frac{2 \sin(\theta) \cos(\theta)}{3} \][/tex]
[tex]\[ \frac{\sin(\theta)}{2 \cos(\theta)} = \frac{2}{3} \][/tex]
[tex]\[ \tan(\theta) = \frac{4}{3} \][/tex]
4. Finding θ:
[tex]\[ \theta = \tan^{-1}\left(\frac{4}{3}\right) \approx 41.41^\circ \][/tex]
#### ii. Time of Flight (T)
1. Equation for Time of Flight:
[tex]\[ T = \frac{2v \sin(\theta)}{g} \][/tex]
2. Substitute values:
[tex]\[ T = \frac{2 \times 250 \times \sin(41.41^\circ)}{10} \][/tex]
[tex]\[ T \approx 33.07 \text{ seconds} \][/tex]
#### iii. Range (R)
1. Equation for Range:
[tex]\[ R = \frac{v^2 \sin(2\theta)}{g} \][/tex]
2. Substitute values:
[tex]\[ R = \frac{250^2 \cdot \sin(82.82^\circ)}{10} \][/tex]
[tex]\[ R \approx 6200.98 \text{ meters} \][/tex]
#### iv. Maximum Height (H)
1. Given relationship:
[tex]\[ H = \frac{R}{3} \][/tex]
2. Substitute range value:
[tex]\[ H = \frac{6200.98}{3} \][/tex]
[tex]\[ H \approx 2066.99 \text{ meters} \][/tex]
### Summary of Results:
1. Angle of Projection (θ): [tex]\(41.41^\circ\)[/tex]
2. Time of Flight (T): [tex]\(33.07\)[/tex] seconds
3. Range (R): [tex]\(6200.98\)[/tex] meters
4. Maximum Height (H): [tex]\(2066.99\)[/tex] meters
### Step-by-Step Solution:
#### Given:
- Initial velocity (v) = 250 m/s
- Acceleration due to gravity (g) = 10 m/s²
- The maximum height (H) is one-third of the range (R)
#### i. Angle at which the bullet is fired (θ)
1. Equations of Projectile Motion:
- Maximum height (H):
[tex]\[ H = \frac{v^2 \sin^2(\theta)}{2g} \][/tex]
- Range (R):
[tex]\[ R = \frac{v^2 \sin(2\theta)}{g} \][/tex]
- Given relationship:
[tex]\[ H = \frac{R}{3} \][/tex]
2. Substituting H = R/3 into equation:
[tex]\[ \frac{v^2 \sin^2(\theta)}{2g} = \frac{1}{3} \left( \frac{v^2 \sin(2\theta)}{g} \right) \][/tex]
Simplifying:
[tex]\[ \frac{\sin^2(\theta)}{2} = \frac{\sin(2\theta)}{3} \][/tex]
3. Using the identity [tex]\(\sin(2\theta) = 2 \sin(\theta) \cos(\theta)\)[/tex]:
[tex]\[ \frac{\sin^2(\theta)}{2} = \frac{2 \sin(\theta) \cos(\theta)}{3} \][/tex]
[tex]\[ \frac{\sin(\theta)}{2 \cos(\theta)} = \frac{2}{3} \][/tex]
[tex]\[ \tan(\theta) = \frac{4}{3} \][/tex]
4. Finding θ:
[tex]\[ \theta = \tan^{-1}\left(\frac{4}{3}\right) \approx 41.41^\circ \][/tex]
#### ii. Time of Flight (T)
1. Equation for Time of Flight:
[tex]\[ T = \frac{2v \sin(\theta)}{g} \][/tex]
2. Substitute values:
[tex]\[ T = \frac{2 \times 250 \times \sin(41.41^\circ)}{10} \][/tex]
[tex]\[ T \approx 33.07 \text{ seconds} \][/tex]
#### iii. Range (R)
1. Equation for Range:
[tex]\[ R = \frac{v^2 \sin(2\theta)}{g} \][/tex]
2. Substitute values:
[tex]\[ R = \frac{250^2 \cdot \sin(82.82^\circ)}{10} \][/tex]
[tex]\[ R \approx 6200.98 \text{ meters} \][/tex]
#### iv. Maximum Height (H)
1. Given relationship:
[tex]\[ H = \frac{R}{3} \][/tex]
2. Substitute range value:
[tex]\[ H = \frac{6200.98}{3} \][/tex]
[tex]\[ H \approx 2066.99 \text{ meters} \][/tex]
### Summary of Results:
1. Angle of Projection (θ): [tex]\(41.41^\circ\)[/tex]
2. Time of Flight (T): [tex]\(33.07\)[/tex] seconds
3. Range (R): [tex]\(6200.98\)[/tex] meters
4. Maximum Height (H): [tex]\(2066.99\)[/tex] meters
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