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Sagot :
Alright, let's break down the problem step by step.
### Understanding the Problem
We are given the height equation for an object thrown upward:
[tex]\[ h = -16t^2 + 79t + 11 \][/tex]
where [tex]\( h \)[/tex] is the height in feet and [tex]\( t \)[/tex] is the time in seconds.
#### Step 1: Finding the time when the height will be 105 feet.
To find the time when the object is at a height of 105 feet, we set [tex]\( h \)[/tex] to 105 and solve for [tex]\( t \)[/tex]:
[tex]\[ -16t^2 + 79t + 11 = 105 \][/tex]
Subtract 105 from both sides to set the equation to 0:
[tex]\[ -16t^2 + 79t + 11 - 105 = 0 \][/tex]
[tex]\[ -16t^2 + 79t - 94 = 0 \][/tex]
Solve this quadratic equation for [tex]\( t \)[/tex]:
The solutions to this quadratic equation are [tex]\( t = 2 \)[/tex] seconds and [tex]\( t = \frac{47}{16} \)[/tex] seconds (or approximately 2.9375 seconds).
So, the object will be at 105 feet at [tex]\( t = 2 \)[/tex] seconds and [tex]\( t = \frac{47}{16} \)[/tex] seconds.
### Summary for 105 feet:
- Time: [tex]\( t = 2 \)[/tex] seconds
- Time: [tex]\( t = \frac{47}{16} \)[/tex] seconds (approximately 2.9375 seconds)
### Step 2: Finding the time when the object reaches the ground.
To find when the object reaches the ground, we set [tex]\( h \)[/tex] to 0 and solve for [tex]\( t \)[/tex]:
[tex]\[ -16t^2 + 79t + 11 = 0 \][/tex]
Solve this quadratic equation for [tex]\( t \)[/tex]:
The solutions to this quadratic equation are:
[tex]\[ t = \frac{79}{32} - \frac{\sqrt{6945}}{32} \][/tex]
[tex]\[ t = \frac{79}{32} + \frac{\sqrt{6945}}{32} \][/tex]
In decimal form, these solutions are approximately:
[tex]\[ t \approx -0.13 \][/tex] (which is not physically meaningful since time cannot be negative)
[tex]\[ t \approx 5.13 \][/tex] seconds
So, the object will reach the ground at [tex]\( t = 5.13 \)[/tex] seconds.
### Summary for reaching the ground:
- Time: [tex]\( t \approx 5.13 \)[/tex] seconds
### Final Answers
1. When will the height be 105 feet?
- The height of 105 feet will be reached at [tex]\( t = 2 \)[/tex] seconds and [tex]\( t = \frac{47}{16} \approx 2.9375 \)[/tex] seconds.
2. When will the object reach the ground?
- The object will reach the ground at [tex]\( t \approx 5.13 \)[/tex] seconds.
### Understanding the Problem
We are given the height equation for an object thrown upward:
[tex]\[ h = -16t^2 + 79t + 11 \][/tex]
where [tex]\( h \)[/tex] is the height in feet and [tex]\( t \)[/tex] is the time in seconds.
#### Step 1: Finding the time when the height will be 105 feet.
To find the time when the object is at a height of 105 feet, we set [tex]\( h \)[/tex] to 105 and solve for [tex]\( t \)[/tex]:
[tex]\[ -16t^2 + 79t + 11 = 105 \][/tex]
Subtract 105 from both sides to set the equation to 0:
[tex]\[ -16t^2 + 79t + 11 - 105 = 0 \][/tex]
[tex]\[ -16t^2 + 79t - 94 = 0 \][/tex]
Solve this quadratic equation for [tex]\( t \)[/tex]:
The solutions to this quadratic equation are [tex]\( t = 2 \)[/tex] seconds and [tex]\( t = \frac{47}{16} \)[/tex] seconds (or approximately 2.9375 seconds).
So, the object will be at 105 feet at [tex]\( t = 2 \)[/tex] seconds and [tex]\( t = \frac{47}{16} \)[/tex] seconds.
### Summary for 105 feet:
- Time: [tex]\( t = 2 \)[/tex] seconds
- Time: [tex]\( t = \frac{47}{16} \)[/tex] seconds (approximately 2.9375 seconds)
### Step 2: Finding the time when the object reaches the ground.
To find when the object reaches the ground, we set [tex]\( h \)[/tex] to 0 and solve for [tex]\( t \)[/tex]:
[tex]\[ -16t^2 + 79t + 11 = 0 \][/tex]
Solve this quadratic equation for [tex]\( t \)[/tex]:
The solutions to this quadratic equation are:
[tex]\[ t = \frac{79}{32} - \frac{\sqrt{6945}}{32} \][/tex]
[tex]\[ t = \frac{79}{32} + \frac{\sqrt{6945}}{32} \][/tex]
In decimal form, these solutions are approximately:
[tex]\[ t \approx -0.13 \][/tex] (which is not physically meaningful since time cannot be negative)
[tex]\[ t \approx 5.13 \][/tex] seconds
So, the object will reach the ground at [tex]\( t = 5.13 \)[/tex] seconds.
### Summary for reaching the ground:
- Time: [tex]\( t \approx 5.13 \)[/tex] seconds
### Final Answers
1. When will the height be 105 feet?
- The height of 105 feet will be reached at [tex]\( t = 2 \)[/tex] seconds and [tex]\( t = \frac{47}{16} \approx 2.9375 \)[/tex] seconds.
2. When will the object reach the ground?
- The object will reach the ground at [tex]\( t \approx 5.13 \)[/tex] seconds.
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