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Answer :
To determine when Jerald is less than 104 feet above the ground, we start with the equation that models his height:
[tex]\[ h = -16t^2 + 729 \][/tex]
We want to find out for what time [tex]\( t \)[/tex] this height [tex]\( h \)[/tex] is less than 104 feet:
[tex]\[ -16t^2 + 729 < 104 \][/tex]
First, solve the inequality for when the height is exactly 104 feet:
[tex]\[ -16t^2 + 729 = 104 \][/tex]
Subtract 104 from both sides:
[tex]\[ -16t^2 + 729 - 104 = 0 \][/tex]
Simplify the equation:
[tex]\[ -16t^2 + 625 = 0 \][/tex]
Add 16t^2 to both sides to isolate t:
[tex]\[ 16t^2 = 625 \][/tex]
Now, divide by 16:
[tex]\[ t^2 = \frac{625}{16} \][/tex]
Taking the square root of both sides:
[tex]\[ t = \pm\sqrt{\frac{625}{16}} \][/tex]
Calculating the square root:
[tex]\[ t = \pm\frac{25}{4} \][/tex]
These are the critical points: [tex]\( t = -\frac{25}{4} \)[/tex] and [tex]\( t = \frac{25}{4} \)[/tex].
Since time [tex]\( t \)[/tex] must be positive (as it represents time starting from the jump), we'll focus on the positive critical point:
[tex]\[ t = \frac{25}{4} = 6.25 \][/tex]
So, Jerald is less than 104 feet above the ground between these critical points. Therefore, the interval of time when Jerald is less than 104 feet above the ground is:
[tex]\[ 0 < t < 6.25 \][/tex]
This aligns with the given option:
[tex]\(0 \leq t \leq 6.25\)[/tex]
[tex]\[ h = -16t^2 + 729 \][/tex]
We want to find out for what time [tex]\( t \)[/tex] this height [tex]\( h \)[/tex] is less than 104 feet:
[tex]\[ -16t^2 + 729 < 104 \][/tex]
First, solve the inequality for when the height is exactly 104 feet:
[tex]\[ -16t^2 + 729 = 104 \][/tex]
Subtract 104 from both sides:
[tex]\[ -16t^2 + 729 - 104 = 0 \][/tex]
Simplify the equation:
[tex]\[ -16t^2 + 625 = 0 \][/tex]
Add 16t^2 to both sides to isolate t:
[tex]\[ 16t^2 = 625 \][/tex]
Now, divide by 16:
[tex]\[ t^2 = \frac{625}{16} \][/tex]
Taking the square root of both sides:
[tex]\[ t = \pm\sqrt{\frac{625}{16}} \][/tex]
Calculating the square root:
[tex]\[ t = \pm\frac{25}{4} \][/tex]
These are the critical points: [tex]\( t = -\frac{25}{4} \)[/tex] and [tex]\( t = \frac{25}{4} \)[/tex].
Since time [tex]\( t \)[/tex] must be positive (as it represents time starting from the jump), we'll focus on the positive critical point:
[tex]\[ t = \frac{25}{4} = 6.25 \][/tex]
So, Jerald is less than 104 feet above the ground between these critical points. Therefore, the interval of time when Jerald is less than 104 feet above the ground is:
[tex]\[ 0 < t < 6.25 \][/tex]
This aligns with the given option:
[tex]\(0 \leq t \leq 6.25\)[/tex]
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