Answer :
To determine the time interval during which Jerald is less than 104 feet above the ground, we need to solve the inequality given by the height equation [tex]\( h = -16t^2 + 729 \)[/tex].
1. Set up the Inequality:
We need to find when Jerald's height [tex]\( h \)[/tex] is less than 104 feet:
[tex]\[
-16t^2 + 729 < 104
\][/tex]
2. Isolate the quadratic term:
Subtract 104 from both sides to simplify the inequality:
[tex]\[
-16t^2 + 729 - 104 < 0
\][/tex]
[tex]\[
-16t^2 + 625 < 0
\][/tex]
3. Simplify the inequality:
Combine like terms:
[tex]\[
-16t^2 + 625 < 0
\][/tex]
4. Solve for [tex]\( t \)[/tex]:
We solve the quadratic inequality by first finding the roots of the equation [tex]\( -16t^2 + 625 = 0 \)[/tex]:
[tex]\[
-16t^2 + 625 = 0
\][/tex]
[tex]\[
-16t^2 = -625
\][/tex]
[tex]\[
t^2 = \frac{625}{16}
\][/tex]
[tex]\[
t^2 = \left(\frac{25}{4}\right)^2
\][/tex]
[tex]\[
t = \pm \frac{25}{4}
\][/tex]
This gives us two critical points: [tex]\( t = \frac{25}{4} \)[/tex] and [tex]\( t = -\frac{25}{4} \)[/tex].
5. Determine the intervals:
The quadratic equation [tex]\( -16t^2 + 625 \)[/tex] changes sign at [tex]\( t = \pm \frac{25}{4} \)[/tex]. To find out the intervals where [tex]\( -16t^2 + 625 < 0 \)[/tex], we test the intervals around these critical points.
Since the quadratic term is negative (leading coefficient [tex]\(-16\)[/tex] is negative) and the parabola opens downwards, the inequality [tex]\( -16t^2 + 625 < 0 \)[/tex] holds outside the interval [tex]\( -\frac{25}{4} < t < \frac{25}{4} \)[/tex].
6. Conclusion:
Observing this result, Jerald is less than 104 feet above the ground when:
[tex]\[
t > \frac{25}{4} \quad \text{or} \quad t < -\frac{25}{4}
\][/tex]
Since [tex]\( -\frac{25}{4} \)[/tex] or [tex]\( -6.25 \)[/tex] seconds doesn’t make sense in a real-world context of time, we only consider the positive part of the interval:
[tex]\[
t > 6.25
\][/tex]
So, the correct interval of time where Jerald is less than 104 feet above the ground is:
[tex]\[
t > 6.25
\][/tex]
1. Set up the Inequality:
We need to find when Jerald's height [tex]\( h \)[/tex] is less than 104 feet:
[tex]\[
-16t^2 + 729 < 104
\][/tex]
2. Isolate the quadratic term:
Subtract 104 from both sides to simplify the inequality:
[tex]\[
-16t^2 + 729 - 104 < 0
\][/tex]
[tex]\[
-16t^2 + 625 < 0
\][/tex]
3. Simplify the inequality:
Combine like terms:
[tex]\[
-16t^2 + 625 < 0
\][/tex]
4. Solve for [tex]\( t \)[/tex]:
We solve the quadratic inequality by first finding the roots of the equation [tex]\( -16t^2 + 625 = 0 \)[/tex]:
[tex]\[
-16t^2 + 625 = 0
\][/tex]
[tex]\[
-16t^2 = -625
\][/tex]
[tex]\[
t^2 = \frac{625}{16}
\][/tex]
[tex]\[
t^2 = \left(\frac{25}{4}\right)^2
\][/tex]
[tex]\[
t = \pm \frac{25}{4}
\][/tex]
This gives us two critical points: [tex]\( t = \frac{25}{4} \)[/tex] and [tex]\( t = -\frac{25}{4} \)[/tex].
5. Determine the intervals:
The quadratic equation [tex]\( -16t^2 + 625 \)[/tex] changes sign at [tex]\( t = \pm \frac{25}{4} \)[/tex]. To find out the intervals where [tex]\( -16t^2 + 625 < 0 \)[/tex], we test the intervals around these critical points.
Since the quadratic term is negative (leading coefficient [tex]\(-16\)[/tex] is negative) and the parabola opens downwards, the inequality [tex]\( -16t^2 + 625 < 0 \)[/tex] holds outside the interval [tex]\( -\frac{25}{4} < t < \frac{25}{4} \)[/tex].
6. Conclusion:
Observing this result, Jerald is less than 104 feet above the ground when:
[tex]\[
t > \frac{25}{4} \quad \text{or} \quad t < -\frac{25}{4}
\][/tex]
Since [tex]\( -\frac{25}{4} \)[/tex] or [tex]\( -6.25 \)[/tex] seconds doesn’t make sense in a real-world context of time, we only consider the positive part of the interval:
[tex]\[
t > 6.25
\][/tex]
So, the correct interval of time where Jerald is less than 104 feet above the ground is:
[tex]\[
t > 6.25
\][/tex]
