To determine how long it will take for the rock to hit the ground, we need to find the time [tex]\( t \)[/tex] when the height [tex]\( h(t) \)[/tex] of the rock is zero. Given the height equation:
[tex]\[ h(t) = -16t^2 + 6.1t + 110 \][/tex]
we set [tex]\( h(t) = 0 \)[/tex] to represent the moment the rock hits the ground:
[tex]\[ -16t^2 + 6.1t + 110 = 0 \][/tex]
This is a standard quadratic equation of the form [tex]\( at^2 + bt + c = 0 \)[/tex], where [tex]\( a = -16 \)[/tex], [tex]\( b = 6.1 \)[/tex], and [tex]\( c = 110 \)[/tex]. To find the solutions for [tex]\( t \)[/tex], we use the quadratic formula:
[tex]\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
First, we compute the discriminant:
[tex]\[ b^2 - 4ac = (6.1)^2 - 4(-16)(110) \][/tex]
[tex]\[ = 37.21 + 7040 \][/tex]
[tex]\[ = 7077.21 \][/tex]
Next, we take the square root of the discriminant:
[tex]\[ \sqrt{7077.21} \approx 84.12 \][/tex]
Now, we can compute the two potential solutions for [tex]\( t \)[/tex]:
[tex]\[ t = \frac{-6.1 \pm 84.12}{-32} \][/tex]
Since we have the plus and minus cases, let's calculate each one:
1. [tex]\( t_1 = \frac{-6.1 + 84.12}{-32} \)[/tex]
[tex]\[ = \frac{78.02}{-32} \][/tex]
[tex]\[ \approx -2.438 \][/tex]
2. [tex]\( t_2 = \frac{-6.1 - 84.12}{-32} \)[/tex]
[tex]\[ = \frac{-90.22}{-32} \][/tex]
[tex]\[ \approx 2.820 \][/tex]
Because time cannot be negative, we discard [tex]\( t_1 \)[/tex]. Thus, we have:
[tex]\[ t = 2.820 \][/tex]
Therefore, it will take approximately 2.820 seconds for the rock to hit the ground.
