Find the best solutions to your questions at Westonci.ca, the premier Q&A platform with a community of knowledgeable experts. Get quick and reliable solutions to your questions from a community of experienced professionals on our platform. Join our Q&A platform to connect with experts dedicated to providing accurate answers to your questions in various fields.
Sagot :
Sure, let's tackle this problem step-by-step!
### Given Conditions:
- Initial height [tex]\( h_0 \)[/tex] = 56.3 feet
- Time to reach highest point [tex]\( t_{\text{apex}} \)[/tex] = 1.2 seconds
- Acceleration due to gravity [tex]\( g \)[/tex] = 32.17 ft/s²
### Part (a): What is the balloon's exact height at [tex]\( t = 2.4 \)[/tex] seconds? Why?
1. Determine the initial velocity [tex]\( v_0 \)[/tex]:
The maximum height is reached when the velocity is 0, which occurs at [tex]\( t_{\text{apex}} \)[/tex]. Using the relation [tex]\( v = v_0 - g \cdot t \)[/tex] (where [tex]\( v \)[/tex] is the velocity),
[tex]\[ 0 = v_0 - g \cdot t_{\text{apex}} \][/tex]
Solving for [tex]\( v_0 \)[/tex]:
[tex]\[ v_0 = g \cdot t_{\text{apex}} = 32.17 \text{ ft/s}^2 \cdot 1.2 \text{ s} = 38.604 \text{ ft/s} \][/tex]
2. Calculate the height at [tex]\( t = 2.4 \)[/tex] seconds:
The height equation is given by [tex]\[ h(t) = h_0 + v_0 \cdot t - \frac{1}{2} g \cdot t^2 \][/tex]
Plugging in the values:
[tex]\[ h(2.4) = 56.3 + 38.604 \cdot 2.4 - \frac{1}{2} \cdot 32.17 \cdot (2.4)^2 \][/tex]
After performing the calculations:
[tex]\[ h(2.4) = 56.3 \][/tex]
3. Conclusion:
So, the balloon's exact height at [tex]\( t = 2.4 \)[/tex] seconds is approximately [tex]\( 56.3 \)[/tex] feet because it returns back to its initial height at this point after climbing and descending.
### Part (b): What is the exact maximum height the balloon reaches at [tex]\( t = 1.2 \)[/tex] seconds?
1. Calculate the height at the maximum point [tex]\( t = 1.2 \)[/tex]:
[tex]\[ h(1.2) = h_0 + v_0 \cdot 1.2 - \frac{1}{2} g \cdot (1.2)^2 \][/tex]
Plugging in the values:
[tex]\[ h(1.2) = 56.3 + 38.604 \cdot 1.2 - \frac{1}{2} \cdot 32.17 \cdot (1.2)^2 \][/tex]
After performing the calculations:
[tex]\[ h(1.2) = 79.4624 \][/tex]
2. Conclusion:
The exact maximum height the balloon reaches at [tex]\( t = 1.2 \)[/tex] seconds is approximately [tex]\( 79.4624 \)[/tex] feet.
### Part (c): What exact time did the balloon land?
1. The balloon lands when the height [tex]\( h(t) = 0 \)[/tex]. We need to solve the height equation for [tex]\( t \)[/tex]:
[tex]\[ h(t) = h_0 + v_0 \cdot t - \frac{1}{2} g \cdot t^2 = 0 \][/tex]
Substituting the known values and solving the quadratic equation:
[tex]\[ 0 = 56.3 + 38.604 \cdot t - \frac{1}{2} \cdot 32.17 \cdot t^2 \][/tex]
Using quadratic solution methods, we find the roots of this equation. The positive root gives the time when the balloon lands:
[tex]\[ t \approx 3.4226 \][/tex]
2. Conclusion:
Therefore, the exact time when the balloon lands is approximately [tex]\( 3.4226 \)[/tex] seconds.
### Part (d): At what initial velocity was the balloon launched?
1. From the calculations in part (a):
[tex]\[ v_0 = g \cdot t_{\text{apex}} = 32.17 \text{ ft/s}^2 \cdot 1.2 \text{ s} = 38.604 \text{ ft/s} \][/tex]
2. Conclusion:
The initial velocity at which the balloon was launched is approximately [tex]\( 38.604 \)[/tex] ft/s.
### Summary:
- Height at [tex]\( t = 2.4 \)[/tex] seconds: [tex]\(56.3\)[/tex] feet
- Maximum height at [tex]\( t = 1.2 \)[/tex] seconds: [tex]\(79.4624\)[/tex] feet
- Balloon lands at: [tex]\(3.4226\)[/tex] seconds
- Initial velocity: [tex]\(38.604\)[/tex] ft/s
These detailed results provide the solution to the problem based on the given initial conditions.
### Given Conditions:
- Initial height [tex]\( h_0 \)[/tex] = 56.3 feet
- Time to reach highest point [tex]\( t_{\text{apex}} \)[/tex] = 1.2 seconds
- Acceleration due to gravity [tex]\( g \)[/tex] = 32.17 ft/s²
### Part (a): What is the balloon's exact height at [tex]\( t = 2.4 \)[/tex] seconds? Why?
1. Determine the initial velocity [tex]\( v_0 \)[/tex]:
The maximum height is reached when the velocity is 0, which occurs at [tex]\( t_{\text{apex}} \)[/tex]. Using the relation [tex]\( v = v_0 - g \cdot t \)[/tex] (where [tex]\( v \)[/tex] is the velocity),
[tex]\[ 0 = v_0 - g \cdot t_{\text{apex}} \][/tex]
Solving for [tex]\( v_0 \)[/tex]:
[tex]\[ v_0 = g \cdot t_{\text{apex}} = 32.17 \text{ ft/s}^2 \cdot 1.2 \text{ s} = 38.604 \text{ ft/s} \][/tex]
2. Calculate the height at [tex]\( t = 2.4 \)[/tex] seconds:
The height equation is given by [tex]\[ h(t) = h_0 + v_0 \cdot t - \frac{1}{2} g \cdot t^2 \][/tex]
Plugging in the values:
[tex]\[ h(2.4) = 56.3 + 38.604 \cdot 2.4 - \frac{1}{2} \cdot 32.17 \cdot (2.4)^2 \][/tex]
After performing the calculations:
[tex]\[ h(2.4) = 56.3 \][/tex]
3. Conclusion:
So, the balloon's exact height at [tex]\( t = 2.4 \)[/tex] seconds is approximately [tex]\( 56.3 \)[/tex] feet because it returns back to its initial height at this point after climbing and descending.
### Part (b): What is the exact maximum height the balloon reaches at [tex]\( t = 1.2 \)[/tex] seconds?
1. Calculate the height at the maximum point [tex]\( t = 1.2 \)[/tex]:
[tex]\[ h(1.2) = h_0 + v_0 \cdot 1.2 - \frac{1}{2} g \cdot (1.2)^2 \][/tex]
Plugging in the values:
[tex]\[ h(1.2) = 56.3 + 38.604 \cdot 1.2 - \frac{1}{2} \cdot 32.17 \cdot (1.2)^2 \][/tex]
After performing the calculations:
[tex]\[ h(1.2) = 79.4624 \][/tex]
2. Conclusion:
The exact maximum height the balloon reaches at [tex]\( t = 1.2 \)[/tex] seconds is approximately [tex]\( 79.4624 \)[/tex] feet.
### Part (c): What exact time did the balloon land?
1. The balloon lands when the height [tex]\( h(t) = 0 \)[/tex]. We need to solve the height equation for [tex]\( t \)[/tex]:
[tex]\[ h(t) = h_0 + v_0 \cdot t - \frac{1}{2} g \cdot t^2 = 0 \][/tex]
Substituting the known values and solving the quadratic equation:
[tex]\[ 0 = 56.3 + 38.604 \cdot t - \frac{1}{2} \cdot 32.17 \cdot t^2 \][/tex]
Using quadratic solution methods, we find the roots of this equation. The positive root gives the time when the balloon lands:
[tex]\[ t \approx 3.4226 \][/tex]
2. Conclusion:
Therefore, the exact time when the balloon lands is approximately [tex]\( 3.4226 \)[/tex] seconds.
### Part (d): At what initial velocity was the balloon launched?
1. From the calculations in part (a):
[tex]\[ v_0 = g \cdot t_{\text{apex}} = 32.17 \text{ ft/s}^2 \cdot 1.2 \text{ s} = 38.604 \text{ ft/s} \][/tex]
2. Conclusion:
The initial velocity at which the balloon was launched is approximately [tex]\( 38.604 \)[/tex] ft/s.
### Summary:
- Height at [tex]\( t = 2.4 \)[/tex] seconds: [tex]\(56.3\)[/tex] feet
- Maximum height at [tex]\( t = 1.2 \)[/tex] seconds: [tex]\(79.4624\)[/tex] feet
- Balloon lands at: [tex]\(3.4226\)[/tex] seconds
- Initial velocity: [tex]\(38.604\)[/tex] ft/s
These detailed results provide the solution to the problem based on the given initial conditions.
Thank you for choosing our service. We're dedicated to providing the best answers for all your questions. Visit us again. We hope you found this helpful. Feel free to come back anytime for more accurate answers and updated information. Keep exploring Westonci.ca for more insightful answers to your questions. We're here to help.
