To answer these questions, we need to solve the given height equation [tex]\( h = -16t^2 + 68t + 9 \)[/tex] for specific conditions.
### Part 1: When will the height be 80 feet?
We need to find the time [tex]\( t \)[/tex] when the height [tex]\( h \)[/tex] is 80 feet:
[tex]\[
80 = -16t^2 + 68t + 9
\][/tex]
Rearrange the equation to set it to zero:
[tex]\[
-16t^2 + 68t + 9 - 80 = 0 \quad \Rightarrow \quad -16t^2 + 68t - 71 = 0
\][/tex]
This is a quadratic equation in the form [tex]\( at^2 + bt + c = 0 \)[/tex]. Here, [tex]\( a = -16 \)[/tex], [tex]\( b = 68 \)[/tex], and [tex]\( c = -71 \)[/tex]. To solve it, we use the quadratic formula:
[tex]\[
t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\][/tex]
First, calculate the discriminant, [tex]\( \Delta \)[/tex]:
[tex]\[
\Delta = b^2 - 4ac = 68^2 - 4(-16)(-71)
\][/tex]
Using the discriminant, we find the two potential solutions for [tex]\( t \)[/tex]:
[tex]\[
t_1 = \frac{-68 + \sqrt{\Delta}}{2 \times -16} \quad \text{and} \quad t_2 = \frac{-68 - \sqrt{\Delta}}{2 \times -16}
\][/tex]
Substituting the values, we get:
[tex]\[
t_1 \approx 1.85 \quad \text{seconds}
\][/tex]
[tex]\[
t_2 \approx 2.40 \quad \text{seconds}
\][/tex]
So, the height of 80 feet is achieved at approximately 1.85 seconds and 2.40 seconds.
[tex]\[
\boxed{1.85 \, \text{seconds}}
\][/tex]
[tex]\[
\boxed{2.40 \, \text{seconds}}
\][/tex]
### Part 2: When will the object reach the ground?
To find when the object will reach the ground, we set [tex]\( h = 0 \)[/tex]:
[tex]\[
0 = -16t^2 + 68t + 9
\][/tex]
This is also a quadratic equation. Here, [tex]\( a = -16 \)[/tex], [tex]\( b = 68 \)[/tex], and [tex]\( c = 9 \)[/tex]. We use the same quadratic formula:
[tex]\[
t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\][/tex]
Calculate the discriminant, [tex]\( \Delta \)[/tex]:
[tex]\[
\Delta = b^2 - 4ac = 68^2 - 4(-16)(9)
\][/tex]
Using the discriminant, find the two potential solutions for [tex]\( t \)[/tex]:
[tex]\[
t_1 = \frac{-68 + \sqrt{\Delta}}{2 \times -16} \quad \text{and} \quad t_2 = \frac{-68 - \sqrt{\Delta}}{2 \times -16}
\][/tex]
Substituting these values, we get:
[tex]\[
t_1 \approx -0.13 \quad \text{seconds (not a valid physical solution)}
\][/tex]
[tex]\[
t_2 \approx 4.38 \quad \text{seconds}
\][/tex]
The valid solution is:
[tex]\[
\boxed{4.38 \, \text{seconds}}
\][/tex]
