Answer :
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.
