To determine how long after the arrow was shot it hits the target placed 20 feet above the ground, we need to solve the quadratic equation derived from the height equation given: [tex]\( h = -16t^2 + 100t + 5 \)[/tex], where [tex]\( h \)[/tex] is the height in feet and [tex]\( t \)[/tex] is the time in seconds.
Given that the target height is 20 feet, we set [tex]\( h = 20 \)[/tex] and solve for [tex]\( t \)[/tex]:
[tex]\[ 20 = -16t^2 + 100t + 5 \][/tex]
First, we rearrange this equation into standard quadratic form:
[tex]\[ -16t^2 + 100t + 5 - 20 = 0 \][/tex]
[tex]\[ -16t^2 + 100t - 15 = 0 \][/tex]
This is a quadratic equation in the form of [tex]\( at^2 + bt + c = 0 \)[/tex], where:
- [tex]\( a = -16 \)[/tex]
- [tex]\( b = 100 \)[/tex]
- [tex]\( c = -15 \)[/tex]
We solve this quadratic equation using the quadratic formula:
[tex]\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Substituting the values for [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex]:
[tex]\[ t = \frac{-100 \pm \sqrt{100^2 - 4(-16)(-15)}}{2(-16)} \][/tex]
This simplifies to:
[tex]\[ t = \frac{-100 \pm \sqrt{10000 - 960}}{-32} \][/tex]
[tex]\[ t = \frac{-100 \pm \sqrt{9040}}{-32} \][/tex]
[tex]\[ t = \frac{-100 \pm 95.096}{-32} \][/tex]
This results in two potential solutions for [tex]\( t \)[/tex]:
[tex]\[ t_1 = \frac{-100 + 95.096}{-32} \][/tex]
[tex]\[ t_1 = \frac{-4.904}{-32} \][/tex]
[tex]\[ t_1 \approx 0.1538 \text{ seconds} \][/tex]
and
[tex]\[ t_2 = \frac{-100 - 95.096}{-32} \][/tex]
[tex]\[ t_2 = \frac{-195.096}{-32} \][/tex]
[tex]\[ t_2 \approx 6.0962 \text{ seconds} \][/tex]
Since time cannot be negative, we choose the positive time solution that makes physical sense. Both [tex]\( t_1 \)[/tex] and [tex]\( t_2 \)[/tex] are positive, but only the larger value occurs after the arrow reaches its peak height and then descends to hit the target.
Therefore, the time when the arrow hits the target is approximately:
[tex]\[ 6.10 \text{ seconds} \][/tex]
Thus, the correct answer is:
6.10 seconds.
