Answer :
To solve the problem of determining when Jerald is less than 104 feet above the ground, we need to analyze the given equation for his height:
[tex]\[ h = -16t^2 + 729 \][/tex]
This equation models Jerald's height [tex]\( h \)[/tex] in feet, where [tex]\( t \)[/tex] is the time in seconds after he jumps.
We want to find when his height, [tex]\( h \)[/tex], is less than 104 feet:
[tex]\[ -16t^2 + 729 < 104 \][/tex]
Let's solve this inequality step-by-step:
1. Subtract 104 from both sides:
[tex]\[ -16t^2 + 729 - 104 < 0 \][/tex]
Simplify the left side:
[tex]\[ -16t^2 + 625 < 0 \][/tex]
2. Rearrange the inequality:
[tex]\[ 625 < 16t^2 \][/tex]
3. Divide both sides by 16 to isolate [tex]\( t^2 \)[/tex]:
[tex]\[ \frac{625}{16} < t^2 \][/tex]
Calculate the left side:
[tex]\[ 39.0625 < t^2 \][/tex]
4. Take the square root of both sides to solve for [tex]\( t \)[/tex]:
[tex]\[ \sqrt{39.0625} < |t| \][/tex]
Calculate the square root:
[tex]\[ 6.25 < |t| \][/tex]
This means that the time [tex]\( t \)[/tex] when Jerald is less than 104 feet above the ground is when the magnitude of [tex]\( t \)[/tex] is greater than 6.25 seconds. In terms of intervals, this implies:
- [tex]\( t < -6.25 \)[/tex] or [tex]\( t > 6.25 \)[/tex]
Since time [tex]\( t \)[/tex] is non-negative (as negative time would not make sense in this context), we consider only the positive interval:
[tex]\[ t > 6.25 \][/tex]
Therefore, Jerald is less than 104 feet above the ground for [tex]\( t > 6.25 \)[/tex] seconds. The correct interval from the options given is:
[tex]\[ t > 6.25 \][/tex]
[tex]\[ h = -16t^2 + 729 \][/tex]
This equation models Jerald's height [tex]\( h \)[/tex] in feet, where [tex]\( t \)[/tex] is the time in seconds after he jumps.
We want to find when his height, [tex]\( h \)[/tex], is less than 104 feet:
[tex]\[ -16t^2 + 729 < 104 \][/tex]
Let's solve this inequality step-by-step:
1. Subtract 104 from both sides:
[tex]\[ -16t^2 + 729 - 104 < 0 \][/tex]
Simplify the left side:
[tex]\[ -16t^2 + 625 < 0 \][/tex]
2. Rearrange the inequality:
[tex]\[ 625 < 16t^2 \][/tex]
3. Divide both sides by 16 to isolate [tex]\( t^2 \)[/tex]:
[tex]\[ \frac{625}{16} < t^2 \][/tex]
Calculate the left side:
[tex]\[ 39.0625 < t^2 \][/tex]
4. Take the square root of both sides to solve for [tex]\( t \)[/tex]:
[tex]\[ \sqrt{39.0625} < |t| \][/tex]
Calculate the square root:
[tex]\[ 6.25 < |t| \][/tex]
This means that the time [tex]\( t \)[/tex] when Jerald is less than 104 feet above the ground is when the magnitude of [tex]\( t \)[/tex] is greater than 6.25 seconds. In terms of intervals, this implies:
- [tex]\( t < -6.25 \)[/tex] or [tex]\( t > 6.25 \)[/tex]
Since time [tex]\( t \)[/tex] is non-negative (as negative time would not make sense in this context), we consider only the positive interval:
[tex]\[ t > 6.25 \][/tex]
Therefore, Jerald is less than 104 feet above the ground for [tex]\( t > 6.25 \)[/tex] seconds. The correct interval from the options given is:
[tex]\[ t > 6.25 \][/tex]
