Answer :
To find the interval of time during which Jerald is less than 104 feet above the ground, we use the given equation for his height: [tex]\( h = -16t^2 + 729 \)[/tex], where [tex]\( t \)[/tex] is the time in seconds.
We want to determine when his height [tex]\( h \)[/tex] is less than 104 feet. Therefore, we set up the inequality:
[tex]\[
-16t^2 + 729 < 104
\][/tex]
First, we solve the inequality:
1. Subtract 104 from both sides:
[tex]\[
-16t^2 + 729 - 104 < 0
\][/tex]
Which simplifies to:
[tex]\[
-16t^2 + 625 < 0
\][/tex]
2. Subtract 729 from both sides to further simplify, letting us isolate the term with [tex]\( t \)[/tex]:
[tex]\[
-16t^2 < -625
\][/tex]
3. Divide both sides by -16, and remember to flip the inequality sign because we are dividing by a negative number:
[tex]\[
t^2 > \frac{625}{16}
\][/tex]
4. Calculate [tex]\(\frac{625}{16}\)[/tex]:
[tex]\[
t^2 > 39.0625
\][/tex]
5. Take the square root of both sides to solve for [tex]\( t \)[/tex]:
[tex]\[
t > \sqrt{39.0625} \quad \text{or} \quad t < -\sqrt{39.0625}
\][/tex]
6. Calculate the square root of 39.0625:
[tex]\[
\sqrt{39.0625} = 6.25
\][/tex]
Therefore, the two solutions for [tex]\( t \)[/tex] are:
[tex]\[
t > 6.25 \quad \text{or} \quad t < -6.25
\][/tex]
Since in the context of this problem, [tex]\( t \)[/tex] represents time and should be non-negative, we discard the negative solution. Thus, Jerald is less than 104 feet above the ground when:
[tex]\[
t > 6.25
\][/tex]
The correct answer is: [tex]\( t > 6.25 \)[/tex].
We want to determine when his height [tex]\( h \)[/tex] is less than 104 feet. Therefore, we set up the inequality:
[tex]\[
-16t^2 + 729 < 104
\][/tex]
First, we solve the inequality:
1. Subtract 104 from both sides:
[tex]\[
-16t^2 + 729 - 104 < 0
\][/tex]
Which simplifies to:
[tex]\[
-16t^2 + 625 < 0
\][/tex]
2. Subtract 729 from both sides to further simplify, letting us isolate the term with [tex]\( t \)[/tex]:
[tex]\[
-16t^2 < -625
\][/tex]
3. Divide both sides by -16, and remember to flip the inequality sign because we are dividing by a negative number:
[tex]\[
t^2 > \frac{625}{16}
\][/tex]
4. Calculate [tex]\(\frac{625}{16}\)[/tex]:
[tex]\[
t^2 > 39.0625
\][/tex]
5. Take the square root of both sides to solve for [tex]\( t \)[/tex]:
[tex]\[
t > \sqrt{39.0625} \quad \text{or} \quad t < -\sqrt{39.0625}
\][/tex]
6. Calculate the square root of 39.0625:
[tex]\[
\sqrt{39.0625} = 6.25
\][/tex]
Therefore, the two solutions for [tex]\( t \)[/tex] are:
[tex]\[
t > 6.25 \quad \text{or} \quad t < -6.25
\][/tex]
Since in the context of this problem, [tex]\( t \)[/tex] represents time and should be non-negative, we discard the negative solution. Thus, Jerald is less than 104 feet above the ground when:
[tex]\[
t > 6.25
\][/tex]
The correct answer is: [tex]\( t > 6.25 \)[/tex].
