Answer :
To find the time interval during which Jerald is less than 104 feet above the ground, we start with the height equation given:
[tex]\[ h = -16t^2 + 729 \][/tex]
We want to determine for which values of [tex]\( t \)[/tex] the height [tex]\( h \)[/tex] is less than 104 feet:
[tex]\[ -16t^2 + 729 < 104 \][/tex]
Step 1: Rearrange the inequality.
Subtract 729 from both sides:
[tex]\[ -16t^2 < 104 - 729 \][/tex]
[tex]\[ -16t^2 < -625 \][/tex]
Step 2: Solve for [tex]\( t^2 \)[/tex].
Divide both sides by -16. Remember that dividing an inequality by a negative number reverses the inequality sign:
[tex]\[ t^2 > \frac{-625}{-16} \][/tex]
[tex]\[ t^2 > 39.0625 \][/tex]
Step 3: Solve for [tex]\( t \)[/tex].
Take the square root of both sides of the inequality:
[tex]\[ t > \sqrt{39.0625} \][/tex]
[tex]\[ t > 6.25 \][/tex]
Since [tex]\( t \)[/tex] represents time, which cannot be negative in this context, we only consider positive values of [tex]\( t \)[/tex].
Thus, Jerald is less than 104 feet above the ground for times greater than [tex]\( 6.25 \)[/tex] seconds.
Therefore, the correct answer is the interval:
[tex]\[ t > 6.25 \][/tex]
[tex]\[ h = -16t^2 + 729 \][/tex]
We want to determine for which values of [tex]\( t \)[/tex] the height [tex]\( h \)[/tex] is less than 104 feet:
[tex]\[ -16t^2 + 729 < 104 \][/tex]
Step 1: Rearrange the inequality.
Subtract 729 from both sides:
[tex]\[ -16t^2 < 104 - 729 \][/tex]
[tex]\[ -16t^2 < -625 \][/tex]
Step 2: Solve for [tex]\( t^2 \)[/tex].
Divide both sides by -16. Remember that dividing an inequality by a negative number reverses the inequality sign:
[tex]\[ t^2 > \frac{-625}{-16} \][/tex]
[tex]\[ t^2 > 39.0625 \][/tex]
Step 3: Solve for [tex]\( t \)[/tex].
Take the square root of both sides of the inequality:
[tex]\[ t > \sqrt{39.0625} \][/tex]
[tex]\[ t > 6.25 \][/tex]
Since [tex]\( t \)[/tex] represents time, which cannot be negative in this context, we only consider positive values of [tex]\( t \)[/tex].
Thus, Jerald is less than 104 feet above the ground for times greater than [tex]\( 6.25 \)[/tex] seconds.
Therefore, the correct answer is the interval:
[tex]\[ t > 6.25 \][/tex]
