To determine when the soccer ball will hit the ground, we need to find the time [tex]\( t \)[/tex] when the height [tex]\( h(t) \)[/tex] is equal to zero. The height of the ball as a function of time is given by:
[tex]\[ h(t) = -16t^2 + 80t + 384 \][/tex]
We need to solve for [tex]\( t \)[/tex] when [tex]\( h(t) = 0 \)[/tex]:
[tex]\[ -16t^2 + 80t + 384 = 0 \][/tex]
This is a quadratic equation of the form [tex]\( at^2 + bt + c = 0 \)[/tex] where [tex]\( a = -16 \)[/tex], [tex]\( b = 80 \)[/tex], and [tex]\( c = 384 \)[/tex].
To solve this quadratic equation, we can use the quadratic formula:
[tex]\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Substituting in the values of [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex]:
[tex]\[ t = \frac{-80 \pm \sqrt{80^2 - 4(-16)(384)}}{2(-16)} \][/tex]
[tex]\[ t = \frac{-80 \pm \sqrt{6400 + 24576}}{-32} \][/tex]
[tex]\[ t = \frac{-80 \pm \sqrt{30976}}{-32} \][/tex]
[tex]\[ t = \frac{-80 \pm 176}{-32} \][/tex]
This gives us two solutions:
1. [tex]\( t = \frac{-80 + 176}{-32} = \frac{96}{-32} = -3 \)[/tex]
2. [tex]\( t = \frac{-80 - 176}{-32} = \frac{-256}{-32} = 8 \)[/tex]
Thus, the two potential solutions for [tex]\( t \)[/tex] are [tex]\( t = -3 \)[/tex] and [tex]\( t = 8 \)[/tex].
Since negative time doesn't make physical sense in this context, we discard [tex]\( t = -3 \)[/tex].
Therefore, the ball will hit the ground at [tex]\( t = 8 \)[/tex] seconds.
