Westonci.ca is the best place to get answers to your questions, provided by a community of experienced and knowledgeable experts. Connect with a community of professionals ready to help you find accurate solutions to your questions quickly and efficiently. Get detailed and accurate answers to your questions from a dedicated community of experts on our Q&A platform.
Sagot :
To solve for the velocity of projection when a stone is projected to achieve a horizontal range of 24 meters and a peak height of 6 meters, we can use principles of projectile motion. Here is the detailed, step-by-step solution for finding the initial velocity [tex]\( U \)[/tex]:
### Step-by-Step Solution:
1. Given Data:
- Horizontal Range ([tex]\( R \)[/tex]) = 24 meters
- Maximum Height ([tex]\( H \)[/tex]) = 6 meters
- Acceleration due to gravity ([tex]\( g \)[/tex]) = 9.81 m/s²
2. Projectile Motion Equations:
For the vertical motion:
[tex]\[ H = \frac{U^2 \sin^2(\theta)}{2g} \][/tex]
For the horizontal motion:
[tex]\[ R = \frac{U^2 \sin(2\theta)}{g} \][/tex]
3. Equation for Maximum Height:
Rearrange the height formula to solve for [tex]\( U \sin(\theta) \)[/tex]:
[tex]\[ H = \frac{U^2 \sin^2(\theta)}{2g} \][/tex]
[tex]\[ U^2 \sin^2(\theta) = 2gH \][/tex]
[tex]\[ U \sin(\theta) = \sqrt{2gH} \][/tex]
4. Equation for Horizontal Range:
Rearrange the range formula to solve for [tex]\( U \cos(\theta) \)[/tex]:
[tex]\[ R = \frac{U^2 \sin(2\theta)}{g} \][/tex]
Knowing that [tex]\(\sin(2\theta) = 2 \sin(\theta) \cos(\theta)\)[/tex]:
[tex]\[ R = \frac{2U^2 \sin(\theta) \cos(\theta)}{g} \][/tex]
[tex]\[ U^2 = \frac{Rg}{2 \sin(\theta) \cos(\theta)} \][/tex]
Since we already calculated [tex]\( U \sin(\theta) \)[/tex]:
[tex]\[ U \cos(\theta) = \frac{R g}{2 \sqrt{2gH}} \][/tex]
5. Calculate Individual Components:
- Calculate [tex]\( U \sin(\theta) \)[/tex]:
[tex]\[ U \sin(\theta) = \sqrt{2 \cdot 9.81 \cdot 6} = \sqrt{117.72} \approx 10.8 \, \text{m/s} \][/tex]
- Calculate [tex]\( U \cos(\theta) \)[/tex]:
[tex]\[ U \cos(\theta) = \frac{24 \cdot 9.81}{2 \cdot \sqrt{2 \cdot 9.81 \cdot 6}} \][/tex]
Substitute the value calculated:
[tex]\[ U \cos(\theta) = \frac{24 \cdot 9.81}{2 \cdot 10.8} \approx 10.9 \, \text{m/s} \][/tex]
6. Resultant Initial Velocity [tex]\( U \)[/tex]:
[tex]\[ U = \sqrt{(U \sin(\theta))^2 + (U \cos(\theta))^2 } \][/tex]
Plug in the values:
[tex]\[ U = \sqrt{10.8^2 + 10.9^2} \][/tex]
[tex]\[ U = \sqrt{116.64 + 118.81} \][/tex]
[tex]\[ U = \sqrt{235.45} \][/tex]
[tex]\[ U \approx 15.3 \, \text{m/s} \][/tex]
Thus, the velocity of projection, [tex]\( U \)[/tex], is approximately [tex]\( 24.3 \, \text{m/s} \)[/tex].
### Step-by-Step Solution:
1. Given Data:
- Horizontal Range ([tex]\( R \)[/tex]) = 24 meters
- Maximum Height ([tex]\( H \)[/tex]) = 6 meters
- Acceleration due to gravity ([tex]\( g \)[/tex]) = 9.81 m/s²
2. Projectile Motion Equations:
For the vertical motion:
[tex]\[ H = \frac{U^2 \sin^2(\theta)}{2g} \][/tex]
For the horizontal motion:
[tex]\[ R = \frac{U^2 \sin(2\theta)}{g} \][/tex]
3. Equation for Maximum Height:
Rearrange the height formula to solve for [tex]\( U \sin(\theta) \)[/tex]:
[tex]\[ H = \frac{U^2 \sin^2(\theta)}{2g} \][/tex]
[tex]\[ U^2 \sin^2(\theta) = 2gH \][/tex]
[tex]\[ U \sin(\theta) = \sqrt{2gH} \][/tex]
4. Equation for Horizontal Range:
Rearrange the range formula to solve for [tex]\( U \cos(\theta) \)[/tex]:
[tex]\[ R = \frac{U^2 \sin(2\theta)}{g} \][/tex]
Knowing that [tex]\(\sin(2\theta) = 2 \sin(\theta) \cos(\theta)\)[/tex]:
[tex]\[ R = \frac{2U^2 \sin(\theta) \cos(\theta)}{g} \][/tex]
[tex]\[ U^2 = \frac{Rg}{2 \sin(\theta) \cos(\theta)} \][/tex]
Since we already calculated [tex]\( U \sin(\theta) \)[/tex]:
[tex]\[ U \cos(\theta) = \frac{R g}{2 \sqrt{2gH}} \][/tex]
5. Calculate Individual Components:
- Calculate [tex]\( U \sin(\theta) \)[/tex]:
[tex]\[ U \sin(\theta) = \sqrt{2 \cdot 9.81 \cdot 6} = \sqrt{117.72} \approx 10.8 \, \text{m/s} \][/tex]
- Calculate [tex]\( U \cos(\theta) \)[/tex]:
[tex]\[ U \cos(\theta) = \frac{24 \cdot 9.81}{2 \cdot \sqrt{2 \cdot 9.81 \cdot 6}} \][/tex]
Substitute the value calculated:
[tex]\[ U \cos(\theta) = \frac{24 \cdot 9.81}{2 \cdot 10.8} \approx 10.9 \, \text{m/s} \][/tex]
6. Resultant Initial Velocity [tex]\( U \)[/tex]:
[tex]\[ U = \sqrt{(U \sin(\theta))^2 + (U \cos(\theta))^2 } \][/tex]
Plug in the values:
[tex]\[ U = \sqrt{10.8^2 + 10.9^2} \][/tex]
[tex]\[ U = \sqrt{116.64 + 118.81} \][/tex]
[tex]\[ U = \sqrt{235.45} \][/tex]
[tex]\[ U \approx 15.3 \, \text{m/s} \][/tex]
Thus, the velocity of projection, [tex]\( U \)[/tex], is approximately [tex]\( 24.3 \, \text{m/s} \)[/tex].
Visit us again for up-to-date and reliable answers. We're always ready to assist you with your informational needs. We hope this was helpful. Please come back whenever you need more information or answers to your queries. We're glad you chose Westonci.ca. Revisit us for updated answers from our knowledgeable team.