Answer :
To solve the problem of finding the time interval during which Jerald is less than 104 feet above the ground, we need to analyze the given height equation:
[tex]\[ h = -16t^2 + 729 \][/tex]
We are looking for the time [tex]\( t \)[/tex] when [tex]\( h < 104 \)[/tex]. So, we start by setting up the inequality:
[tex]\[ -16t^2 + 729 < 104 \][/tex]
Next, we need to solve this inequality step-by-step:
1. Subtract 104 from both sides to move all terms to one side of the inequality:
[tex]\[ -16t^2 + 729 - 104 < 0 \][/tex]
[tex]\[ -16t^2 + 625 < 0 \][/tex]
2. Rearrange the inequality:
[tex]\[ 625 - 16t^2 < 0 \][/tex]
3. Rewrite it to isolate the quadratic term:
[tex]\[ -16t^2 < -625 \][/tex]
4. Since dividing or multiplying by a negative number reverses the inequality sign:
[tex]\[ 16t^2 > 625 \][/tex]
5. Divide both sides by 16 to solve for [tex]\( t^2 \)[/tex]:
[tex]\[ t^2 > \frac{625}{16} \][/tex]
[tex]\[ t^2 > 39.0625 \][/tex]
6. Solve for [tex]\( t \)[/tex] by taking the square root of both sides:
[tex]\[ t > \sqrt{39.0625} \][/tex]
[tex]\[ t > 6.25 \][/tex]
and
[tex]\[ t < -\sqrt{39.0625} \][/tex]
[tex]\[ t < -6.25 \][/tex]
However, negative time does not make sense in this context, so we discard [tex]\( t < -6.25 \)[/tex].
Therefore, we conclude that Jerald is less than 104 feet above the ground when:
[tex]\[ t > 6.25 \][/tex]
So the correct interval of time is:
[tex]\[ t > 6.25 \][/tex]
[tex]\[ h = -16t^2 + 729 \][/tex]
We are looking for the time [tex]\( t \)[/tex] when [tex]\( h < 104 \)[/tex]. So, we start by setting up the inequality:
[tex]\[ -16t^2 + 729 < 104 \][/tex]
Next, we need to solve this inequality step-by-step:
1. Subtract 104 from both sides to move all terms to one side of the inequality:
[tex]\[ -16t^2 + 729 - 104 < 0 \][/tex]
[tex]\[ -16t^2 + 625 < 0 \][/tex]
2. Rearrange the inequality:
[tex]\[ 625 - 16t^2 < 0 \][/tex]
3. Rewrite it to isolate the quadratic term:
[tex]\[ -16t^2 < -625 \][/tex]
4. Since dividing or multiplying by a negative number reverses the inequality sign:
[tex]\[ 16t^2 > 625 \][/tex]
5. Divide both sides by 16 to solve for [tex]\( t^2 \)[/tex]:
[tex]\[ t^2 > \frac{625}{16} \][/tex]
[tex]\[ t^2 > 39.0625 \][/tex]
6. Solve for [tex]\( t \)[/tex] by taking the square root of both sides:
[tex]\[ t > \sqrt{39.0625} \][/tex]
[tex]\[ t > 6.25 \][/tex]
and
[tex]\[ t < -\sqrt{39.0625} \][/tex]
[tex]\[ t < -6.25 \][/tex]
However, negative time does not make sense in this context, so we discard [tex]\( t < -6.25 \)[/tex].
Therefore, we conclude that Jerald is less than 104 feet above the ground when:
[tex]\[ t > 6.25 \][/tex]
So the correct interval of time is:
[tex]\[ t > 6.25 \][/tex]
