Answer :
To find out for which interval of time Jerald is less than 104 feet above the ground, we will use the equation modeling his height:
[tex]\[ h = -16t^2 + 729 \][/tex]
We need to determine the time when his height [tex]\( h \)[/tex] is less than 104 feet. Let's solve the inequality:
[tex]\[ -16t^2 + 729 < 104 \][/tex]
First, we move 104 to the other side of the inequality:
[tex]\[ -16t^2 + 729 - 104 < 0 \][/tex]
[tex]\[ -16t^2 + 625 < 0 \][/tex]
Next, we solve for [tex]\( t \)[/tex]:
1. Set the equation [tex]\( -16t^2 + 625 = 0 \)[/tex] to find the boundary points.
2. Solve for [tex]\( t \)[/tex]:
[tex]\[ -16t^2 + 625 = 0 \][/tex]
[tex]\[ 16t^2 = 625 \][/tex]
[tex]\[ t^2 = \frac{625}{16} \][/tex]
[tex]\[ t^2 = \left(\frac{25}{4}\right)^2 \][/tex]
[tex]\[ t = \pm \frac{25}{4} \][/tex]
[tex]\[ t = \pm 6.25 \][/tex]
3. So, the critical points are [tex]\( t = -6.25 \)[/tex] and [tex]\( t = 6.25 \)[/tex].
Since time [tex]\( t \)[/tex] cannot be negative in this context (Jerald jumped from the bungee tower and [tex]\( t \)[/tex] represents the time after he jumped), we only consider the positive interval. The less than condition applies between these critical points.
Thus, solving [tex]\( -16t^2 + 729 < 104 \)[/tex] and considering realistic time values, we find:
[tex]\[ 0 \leq t \leq 6.25 \][/tex]
Therefore, Jerald is less than 104 feet above the ground for the interval:
[tex]\[ 0 \leq t \leq 6.25 \][/tex]
So the correct answer is:
[tex]\[ 0 \leq t \leq 6.25 \][/tex]
[tex]\[ h = -16t^2 + 729 \][/tex]
We need to determine the time when his height [tex]\( h \)[/tex] is less than 104 feet. Let's solve the inequality:
[tex]\[ -16t^2 + 729 < 104 \][/tex]
First, we move 104 to the other side of the inequality:
[tex]\[ -16t^2 + 729 - 104 < 0 \][/tex]
[tex]\[ -16t^2 + 625 < 0 \][/tex]
Next, we solve for [tex]\( t \)[/tex]:
1. Set the equation [tex]\( -16t^2 + 625 = 0 \)[/tex] to find the boundary points.
2. Solve for [tex]\( t \)[/tex]:
[tex]\[ -16t^2 + 625 = 0 \][/tex]
[tex]\[ 16t^2 = 625 \][/tex]
[tex]\[ t^2 = \frac{625}{16} \][/tex]
[tex]\[ t^2 = \left(\frac{25}{4}\right)^2 \][/tex]
[tex]\[ t = \pm \frac{25}{4} \][/tex]
[tex]\[ t = \pm 6.25 \][/tex]
3. So, the critical points are [tex]\( t = -6.25 \)[/tex] and [tex]\( t = 6.25 \)[/tex].
Since time [tex]\( t \)[/tex] cannot be negative in this context (Jerald jumped from the bungee tower and [tex]\( t \)[/tex] represents the time after he jumped), we only consider the positive interval. The less than condition applies between these critical points.
Thus, solving [tex]\( -16t^2 + 729 < 104 \)[/tex] and considering realistic time values, we find:
[tex]\[ 0 \leq t \leq 6.25 \][/tex]
Therefore, Jerald is less than 104 feet above the ground for the interval:
[tex]\[ 0 \leq t \leq 6.25 \][/tex]
So the correct answer is:
[tex]\[ 0 \leq t \leq 6.25 \][/tex]
