College

Jerald jumped from a bungee tower. If 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]

Answer :

Let's solve the problem of determining the time interval during which Jerald is less than 104 feet above the ground after jumping from the bungee tower. The height of Jerald above the ground over time is modeled by the equation [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 find when Jerald's height, [tex]\( h \)[/tex], is less than 104 feet. This means solving the inequality:

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

1. Rearrange the inequality:

Subtract 729 from both sides:

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

Simplifying the right side:

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

2. Divide by [tex]\(-16\)[/tex]:

Since we are dividing an inequality by a negative number, we reverse the inequality sign:

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

3. Solve for [tex]\(t\)[/tex]:

Calculate [tex]\( \frac{625}{16} \)[/tex]:

[tex]\[ t^2 > 39.0625 \][/tex]

Take the square root of both sides:

[tex]\[ t > \sqrt{39.0625} \quad \text{or} \quad t < -\sqrt{39.0625} \][/tex]

Which gives approximately:

[tex]\[ t > 6.25 \quad \text{or} \quad t < -6.25 \][/tex]

Since negative time doesn't apply in this context of the problem (Jerald jumping at [tex]\(t = 0\)[/tex] and moving forward), we only consider positive [tex]\(t\)[/tex]:

4. Determine the relevant interval:

For practical purposes, within the context of time after the jump starts, Jerald is less than 104 feet above the ground when [tex]\( t > 6.25 \)[/tex].

Therefore, Jerald's height is less than 104 feet for [tex]\( t > 6.25 \)[/tex] seconds.