Answer :
Let's solve the given problem step by step.
### Part 1: When will the height be 151 feet?
We start with the height equation:
[tex]\[ h = -16t^2 + 198t + 6 \][/tex]
We need to find the time \( t \) when the height \( h \) is 151 feet. So we set \( h = 151 \):
[tex]\[ 151 = -16t^2 + 198t + 6 \][/tex]
Rewriting it in standard form (ax^2 + bx + c = 0) gives:
[tex]\[ -16t^2 + 198t + 6 - 151 = 0 \][/tex]
[tex]\[ -16t^2 + 198t - 145 = 0 \][/tex]
This is a quadratic equation of the form \( at^2 + bt + c = 0 \). Here, \( a = -16 \), \( b = 198 \), and \( c = -145 \).
Quadratic equations can be solved using the quadratic formula:
[tex]\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Substitute the values \( a = -16 \), \( b = 198 \), and \( c = -145 \):
[tex]\[ t = \frac{-198 \pm \sqrt{198^2 - 4(-16)(-145)}}{2(-16)} \][/tex]
Calculating the discriminant:
[tex]\[ \sqrt{198^2 - 4(-16)(-145)} = \sqrt{39204 - 9280} = \sqrt{29924} \][/tex]
Therefore, the times are:
[tex]\[ t = \frac{-198 \pm \sqrt{29924}}{-32} \][/tex]
[tex]\[ t = \frac{-198 \pm \sqrt{7481}}{-16} \][/tex]
Simplifying further:
[tex]\[ t = \frac{198}{32} \pm \frac{\sqrt{7481}}{32} \][/tex]
This results in two solutions for \( t \):
[tex]\[ t_1 = \frac{99}{16} - \frac{\sqrt{7481}}{16} \][/tex]
[tex]\[ t_2 = \frac{99}{16} + \frac{\sqrt{7481}}{16} \][/tex]
So, the object will be at a height of 151 feet at:
- \( t_1 = \frac{99}{16} - \frac{\sqrt{7481}}{16} \)
- \( t_2 = \frac{99}{16} + \frac{\sqrt{7481}}{16} \)
### Part 2: When will the object reach the ground?
The object reaches the ground when the height \( h \) is 0. So, we set \( h = 0 \) in the height equation:
[tex]\[ 0 = -16t^2 + 198t + 6 \][/tex]
Again, this is a quadratic equation of the form \( at^2 + bt + c = 0 \). Here, \( a = -16 \), \( b = 198 \), and \( c = 6 \).
Using the quadratic formula we have:
[tex]\[ t = \frac{-198 \pm \sqrt{198^2 - 4(-16)(6)}}{2(-16)} \][/tex]
Calculating the discriminant:
[tex]\[ \sqrt{198^2 - 4(-16)(6)} = \sqrt{39204 + 384} = \sqrt{39588} = \sqrt{9897} \][/tex]
Therefore, the times are:
[tex]\[ t = \frac{-198 \pm \sqrt{9897}}{-32} \][/tex]
[tex]\[ t = \frac{-198 \pm \sqrt{39588}}{-32} \][/tex]
Simplifying further:
[tex]\[ t = \frac{198}{32} \pm \frac{\sqrt{39588}}{32} \][/tex]
This results in two solutions for \( t \):
[tex]\[ t_1 = \frac{99}{16} - \frac{\sqrt{9897}}{16} \][/tex]
[tex]\[ t_2 = \frac{99}{16} + \frac{\sqrt{9897}}{16} \][/tex]
So, the object will reach the ground at:
- \( t_1 = \frac{99}{16} - \frac{\sqrt{9897}}{16} \)
- \( t_2 = \frac{99}{16} + \frac{\sqrt{9897}}{16} \)
Let's summarize the results:
- When will the height be 151 feet?
- At \( t = \frac{99}{16} - \frac{\sqrt{7481}}{16} \)
- And at \( t = \frac{99}{16} + \frac{\sqrt{7481}}{16} \)
- When will the object reach the ground?
- At \( t = \frac{99}{16} - \frac{\sqrt{9897}}{16} \)
- And at [tex]\( t = \frac{99}{16} + \frac{\sqrt{9897}}{16} \)[/tex]
### Part 1: When will the height be 151 feet?
We start with the height equation:
[tex]\[ h = -16t^2 + 198t + 6 \][/tex]
We need to find the time \( t \) when the height \( h \) is 151 feet. So we set \( h = 151 \):
[tex]\[ 151 = -16t^2 + 198t + 6 \][/tex]
Rewriting it in standard form (ax^2 + bx + c = 0) gives:
[tex]\[ -16t^2 + 198t + 6 - 151 = 0 \][/tex]
[tex]\[ -16t^2 + 198t - 145 = 0 \][/tex]
This is a quadratic equation of the form \( at^2 + bt + c = 0 \). Here, \( a = -16 \), \( b = 198 \), and \( c = -145 \).
Quadratic equations can be solved using the quadratic formula:
[tex]\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Substitute the values \( a = -16 \), \( b = 198 \), and \( c = -145 \):
[tex]\[ t = \frac{-198 \pm \sqrt{198^2 - 4(-16)(-145)}}{2(-16)} \][/tex]
Calculating the discriminant:
[tex]\[ \sqrt{198^2 - 4(-16)(-145)} = \sqrt{39204 - 9280} = \sqrt{29924} \][/tex]
Therefore, the times are:
[tex]\[ t = \frac{-198 \pm \sqrt{29924}}{-32} \][/tex]
[tex]\[ t = \frac{-198 \pm \sqrt{7481}}{-16} \][/tex]
Simplifying further:
[tex]\[ t = \frac{198}{32} \pm \frac{\sqrt{7481}}{32} \][/tex]
This results in two solutions for \( t \):
[tex]\[ t_1 = \frac{99}{16} - \frac{\sqrt{7481}}{16} \][/tex]
[tex]\[ t_2 = \frac{99}{16} + \frac{\sqrt{7481}}{16} \][/tex]
So, the object will be at a height of 151 feet at:
- \( t_1 = \frac{99}{16} - \frac{\sqrt{7481}}{16} \)
- \( t_2 = \frac{99}{16} + \frac{\sqrt{7481}}{16} \)
### Part 2: When will the object reach the ground?
The object reaches the ground when the height \( h \) is 0. So, we set \( h = 0 \) in the height equation:
[tex]\[ 0 = -16t^2 + 198t + 6 \][/tex]
Again, this is a quadratic equation of the form \( at^2 + bt + c = 0 \). Here, \( a = -16 \), \( b = 198 \), and \( c = 6 \).
Using the quadratic formula we have:
[tex]\[ t = \frac{-198 \pm \sqrt{198^2 - 4(-16)(6)}}{2(-16)} \][/tex]
Calculating the discriminant:
[tex]\[ \sqrt{198^2 - 4(-16)(6)} = \sqrt{39204 + 384} = \sqrt{39588} = \sqrt{9897} \][/tex]
Therefore, the times are:
[tex]\[ t = \frac{-198 \pm \sqrt{9897}}{-32} \][/tex]
[tex]\[ t = \frac{-198 \pm \sqrt{39588}}{-32} \][/tex]
Simplifying further:
[tex]\[ t = \frac{198}{32} \pm \frac{\sqrt{39588}}{32} \][/tex]
This results in two solutions for \( t \):
[tex]\[ t_1 = \frac{99}{16} - \frac{\sqrt{9897}}{16} \][/tex]
[tex]\[ t_2 = \frac{99}{16} + \frac{\sqrt{9897}}{16} \][/tex]
So, the object will reach the ground at:
- \( t_1 = \frac{99}{16} - \frac{\sqrt{9897}}{16} \)
- \( t_2 = \frac{99}{16} + \frac{\sqrt{9897}}{16} \)
Let's summarize the results:
- When will the height be 151 feet?
- At \( t = \frac{99}{16} - \frac{\sqrt{7481}}{16} \)
- And at \( t = \frac{99}{16} + \frac{\sqrt{7481}}{16} \)
- When will the object reach the ground?
- At \( t = \frac{99}{16} - \frac{\sqrt{9897}}{16} \)
- And at [tex]\( t = \frac{99}{16} + \frac{\sqrt{9897}}{16} \)[/tex]
