Answer :
To determine for which interval of time Jerald is less than 104 feet above the ground, let's solve the inequality given by the height equation:
[tex]\[ h = -16t^2 + 729 < 104 \][/tex]
1. Start by isolating the inequality:
[tex]\[ -16t^2 + 729 < 104 \][/tex]
2. Subtract 104 from both sides to set the inequality to zero:
[tex]\[ -16t^2 + 729 - 104 < 0 \][/tex]
[tex]\[ -16t^2 + 625 < 0 \][/tex]
3. Simplify the equation further:
[tex]\[ -16t^2 + 625 < 0 \][/tex]
4. To solve this inequality, set the quadratic expression to equal zero and solve for [tex]\( t \)[/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]
[tex]\[ t = \pm 6.25 \][/tex]
5. Now, we have the critical points where the height is exactly 104 feet: [tex]\( t = 6.25 \)[/tex] and [tex]\( t = -6.25 \)[/tex]. These points divide the timeline into intervals.
6. We need to determine the intervals around these points where the original inequality is satisfied. The intervals are:
[tex]\[ t < -6.25 \][/tex]
[tex]\[ -6.25 < t < 6.25 \][/tex]
[tex]\[ t > 6.25 \][/tex]
7. To determine where the height is less than 104 feet, we observe that [tex]\( -16t^2 + 625 \)[/tex] is a downward-opening parabola, meaning it will be negative between the roots and positive outside. Therefore:
[tex]\[ 104 > h = -16t^2 + 729 \text{ for } t < -6.25 \text{ and } t > 6.25 \][/tex]
By matching the correct intervals, the solution indicates:
[tex]\[ t > 6.25 \][/tex]
Therefore, the interval of time for which Jerald is less than 104 feet above the ground is [tex]\( t > 6.25 \)[/tex].
[tex]\[ h = -16t^2 + 729 < 104 \][/tex]
1. Start by isolating the inequality:
[tex]\[ -16t^2 + 729 < 104 \][/tex]
2. Subtract 104 from both sides to set the inequality to zero:
[tex]\[ -16t^2 + 729 - 104 < 0 \][/tex]
[tex]\[ -16t^2 + 625 < 0 \][/tex]
3. Simplify the equation further:
[tex]\[ -16t^2 + 625 < 0 \][/tex]
4. To solve this inequality, set the quadratic expression to equal zero and solve for [tex]\( t \)[/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]
[tex]\[ t = \pm 6.25 \][/tex]
5. Now, we have the critical points where the height is exactly 104 feet: [tex]\( t = 6.25 \)[/tex] and [tex]\( t = -6.25 \)[/tex]. These points divide the timeline into intervals.
6. We need to determine the intervals around these points where the original inequality is satisfied. The intervals are:
[tex]\[ t < -6.25 \][/tex]
[tex]\[ -6.25 < t < 6.25 \][/tex]
[tex]\[ t > 6.25 \][/tex]
7. To determine where the height is less than 104 feet, we observe that [tex]\( -16t^2 + 625 \)[/tex] is a downward-opening parabola, meaning it will be negative between the roots and positive outside. Therefore:
[tex]\[ 104 > h = -16t^2 + 729 \text{ for } t < -6.25 \text{ and } t > 6.25 \][/tex]
By matching the correct intervals, the solution indicates:
[tex]\[ t > 6.25 \][/tex]
Therefore, the interval of time for which Jerald is less than 104 feet above the ground is [tex]\( t > 6.25 \)[/tex].
