Answer :
To solve the problem of determining when Jerald's height is less than 104 feet, we start with the given equation for his height:
[tex]\[ h = -16t^2 + 729 \][/tex]
We need to find the interval of time [tex]\( t \)[/tex] for which Jerald's height [tex]\( h \)[/tex] is less than 104 feet:
[tex]\[ -16t^2 + 729 < 104 \][/tex]
First, we will rearrange the inequality to isolate the quadratic expression:
[tex]\[ -16t^2 + 729 - 104 < 0 \][/tex]
Simplifying the equation, we get:
[tex]\[ -16t^2 + 625 < 0 \][/tex]
Now, let's solve this inequality:
1. Quadratic equation: Start by setting up the equation for when it equals zero:
[tex]\[ -16t^2 + 625 = 0 \][/tex]
2. Isolate the quadratic term:
[tex]\[ -16t^2 = -625 \][/tex]
3. Divide both sides by -16:
[tex]\[ t^2 = \frac{625}{16} \][/tex]
4. Take the square root of both sides to solve for [tex]\( t \)[/tex]:
[tex]\[ t = \pm \sqrt{\frac{625}{16}} \][/tex]
5. Calculate the square root:
[tex]\[ t = \pm \frac{25}{4} \][/tex]
This gives us two values: [tex]\( t = \frac{25}{4} \)[/tex] and [tex]\( t = -\frac{25}{4} \)[/tex].
Next, determine the interval where the original inequality holds:
- The values [tex]\( t = \frac{25}{4} \)[/tex] and [tex]\( t = -\frac{25}{4} \)[/tex] are the points where Jerald's height is exactly 104 feet, which is the boundary of our inequality.
- For [tex]\( h \)[/tex] to be less than 104 feet, the time [tex]\( t \)[/tex] needs to be:
[tex]\((-\infty, -\frac{25}{4}) \cup (\frac{25}{4}, \infty)\)[/tex]
Since time cannot be negative in this context (Jerald starts at [tex]\( t = 0 \)[/tex]), we consider only positive [tex]\( t \)[/tex]:
Thus, the interval of time for which Jerald is less than 104 feet above the ground is:
[tex]\[ t > \frac{25}{4} \][/tex]
Converting [tex]\(\frac{25}{4}\)[/tex] to a decimal gives approximately [tex]\(6.25\)[/tex]. Therefore, Jerald is less than 104 feet above the ground for [tex]\( t > 6.25 \)[/tex].
The correct answer from the options provided is:
[tex]\( t > 6.25 \)[/tex]
[tex]\[ h = -16t^2 + 729 \][/tex]
We need to find the interval of time [tex]\( t \)[/tex] for which Jerald's height [tex]\( h \)[/tex] is less than 104 feet:
[tex]\[ -16t^2 + 729 < 104 \][/tex]
First, we will rearrange the inequality to isolate the quadratic expression:
[tex]\[ -16t^2 + 729 - 104 < 0 \][/tex]
Simplifying the equation, we get:
[tex]\[ -16t^2 + 625 < 0 \][/tex]
Now, let's solve this inequality:
1. Quadratic equation: Start by setting up the equation for when it equals zero:
[tex]\[ -16t^2 + 625 = 0 \][/tex]
2. Isolate the quadratic term:
[tex]\[ -16t^2 = -625 \][/tex]
3. Divide both sides by -16:
[tex]\[ t^2 = \frac{625}{16} \][/tex]
4. Take the square root of both sides to solve for [tex]\( t \)[/tex]:
[tex]\[ t = \pm \sqrt{\frac{625}{16}} \][/tex]
5. Calculate the square root:
[tex]\[ t = \pm \frac{25}{4} \][/tex]
This gives us two values: [tex]\( t = \frac{25}{4} \)[/tex] and [tex]\( t = -\frac{25}{4} \)[/tex].
Next, determine the interval where the original inequality holds:
- The values [tex]\( t = \frac{25}{4} \)[/tex] and [tex]\( t = -\frac{25}{4} \)[/tex] are the points where Jerald's height is exactly 104 feet, which is the boundary of our inequality.
- For [tex]\( h \)[/tex] to be less than 104 feet, the time [tex]\( t \)[/tex] needs to be:
[tex]\((-\infty, -\frac{25}{4}) \cup (\frac{25}{4}, \infty)\)[/tex]
Since time cannot be negative in this context (Jerald starts at [tex]\( t = 0 \)[/tex]), we consider only positive [tex]\( t \)[/tex]:
Thus, the interval of time for which Jerald is less than 104 feet above the ground is:
[tex]\[ t > \frac{25}{4} \][/tex]
Converting [tex]\(\frac{25}{4}\)[/tex] to a decimal gives approximately [tex]\(6.25\)[/tex]. Therefore, Jerald is less than 104 feet above the ground for [tex]\( t > 6.25 \)[/tex].
The correct answer from the options provided is:
[tex]\( t > 6.25 \)[/tex]
