Answer :
Let's solve the problem and determine the time interval where Jerald is less than 104 feet above the ground.
The equation given for Jerald's height is:
[tex]\[ h = -16t^2 + 729 \][/tex]
where [tex]\( h \)[/tex] is Jerald's height in feet and [tex]\( t \)[/tex] is the time in seconds.
We want to find the time intervals when Jerald's height is less than 104 feet. So, we set up the inequality:
[tex]\[ -16t^2 + 729 < 104 \][/tex]
First, solve this inequality for [tex]\( t \)[/tex].
1. Subtract 104 from both sides:
[tex]\[ -16t^2 + 729 - 104 < 0 \][/tex]
2. Simplify the equation:
[tex]\[ -16t^2 + 625 < 0 \][/tex]
3. Rearrange the terms:
[tex]\[ 16t^2 > 625 \][/tex]
4. Divide by 16:
[tex]\[ t^2 > \frac{625}{16} \][/tex]
5. Calculate [tex]\( \frac{625}{16} \)[/tex]:
[tex]\[ t^2 > 39.0625 \][/tex]
6. 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]
7. Calculate the square root:
[tex]\[ t > 6.25 \quad \text{or} \quad t < -6.25 \][/tex]
Since time ([tex]\( t \)[/tex]) cannot be a negative value (because Jerald couldn't have jumped before [tex]\( t = 0 \)[/tex]), we consider only positive [tex]\( t \)[/tex]. So, the solution is:
[tex]\[ t > 6.25 \][/tex]
Therefore, the interval for which Jerald is less than 104 feet above the ground is: [tex]\( t > 6.25 \)[/tex].
The equation given for Jerald's height is:
[tex]\[ h = -16t^2 + 729 \][/tex]
where [tex]\( h \)[/tex] is Jerald's height in feet and [tex]\( t \)[/tex] is the time in seconds.
We want to find the time intervals when Jerald's height is less than 104 feet. So, we set up the inequality:
[tex]\[ -16t^2 + 729 < 104 \][/tex]
First, solve this inequality for [tex]\( t \)[/tex].
1. Subtract 104 from both sides:
[tex]\[ -16t^2 + 729 - 104 < 0 \][/tex]
2. Simplify the equation:
[tex]\[ -16t^2 + 625 < 0 \][/tex]
3. Rearrange the terms:
[tex]\[ 16t^2 > 625 \][/tex]
4. Divide by 16:
[tex]\[ t^2 > \frac{625}{16} \][/tex]
5. Calculate [tex]\( \frac{625}{16} \)[/tex]:
[tex]\[ t^2 > 39.0625 \][/tex]
6. 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]
7. Calculate the square root:
[tex]\[ t > 6.25 \quad \text{or} \quad t < -6.25 \][/tex]
Since time ([tex]\( t \)[/tex]) cannot be a negative value (because Jerald couldn't have jumped before [tex]\( t = 0 \)[/tex]), we consider only positive [tex]\( t \)[/tex]. So, the solution is:
[tex]\[ t > 6.25 \][/tex]
Therefore, the interval for which Jerald is less than 104 feet above the ground is: [tex]\( t > 6.25 \)[/tex].
