Jerald jumped from a bungee tower. The equation that models his height, in feet, is [tex]h = -16t^2 + 729[/tex], where [tex]t[/tex] is the time in seconds.

For which interval of time is he less than 104 feet above the ground?

A. [tex]t \ \textgreater \ 6.25[/tex]
B. [tex]-6.25 \ \textless \ t \ \textless \ 6.25[/tex]
C. [tex]t \ \textless \ 6.25[/tex]
D. [tex]0 \leq t \leq 6.25[/tex]

To solve the problem of finding when Jerald is less than 104 feet above the ground, we start with the height equation given by [tex]\( h = -16t^2 + 729 \)[/tex], where [tex]\( h \)[/tex] is the height in feet, and [tex]\( t \)[/tex] is the time in seconds.

We need to determine when:

[tex]\[ -16t^2 + 729 < 104 \][/tex]

Let's break it down step-by-step:

1. Rearrange the inequality:

[tex]\[
-16t^2 + 729 - 104 < 0
\][/tex]

Simplify the expression:

[tex]\[
-16t^2 + 625 < 0
\][/tex]

2. Solve the inequality [tex]\( -16t^2 + 625 < 0 \)[/tex]:

Rearrange it to find when:

[tex]\[
16t^2 > 625
\][/tex]

3. Divide both sides by 16 to isolate [tex]\( t^2 \)[/tex]:

[tex]\[
t^2 > \frac{625}{16}
\][/tex]

4. Calculate the square root of both sides to find [tex]\( t \)[/tex]:

[tex]\[
t > \sqrt{\frac{625}{16}} \quad \text{or} \quad t < -\sqrt{\frac{625}{16}}
\][/tex]

5. Simplify the square root:

[tex]\[
\sqrt{\frac{625}{16}} = \frac{\sqrt{625}}{\sqrt{16}} = \frac{25}{4} = 6.25
\][/tex]

Based on this, the critical points are at [tex]\( t = 6.25 \)[/tex] seconds. Since we are considering a practical situation where time cannot be negative (you can't have negative seconds when Jerald is jumping), we exclude the negative interval.

Therefore, Jerald is less than 104 feet above the ground when:

[tex]\[ t > 6.25 \][/tex]

Thus, the correct choice is [tex]\( t > 6.25 \)[/tex].