Sure, let's solve this problem step-by-step.
First, we start with the given equation for the distance [tex]\(d\)[/tex] from the ground:
[tex]\[ d = 180 + 4t - 0.5 \cdot 9.8 \cdot t^2 \][/tex]
Simplifying the equation, we get:
[tex]\[ d = 180 + 4t - 4.9t^2 \][/tex]
We need to find the time [tex]\(t\)[/tex] when the rocket strikes the ground. When the rocket is on the ground, [tex]\(d = 0\)[/tex]. Therefore, we set the equation to 0:
[tex]\[ 0 = 180 + 4t - 4.9t^2 \][/tex]
Rearranging the terms to get it in the standard quadratic equation form [tex]\(at^2 + bt + c = 0\)[/tex]:
[tex]\[ -4.9t^2 + 4t + 180 = 0 \][/tex]
Here, [tex]\(a = -4.9\)[/tex], [tex]\(b = 4\)[/tex], and [tex]\(c = 180\)[/tex].
To solve this quadratic equation, we use the quadratic formula [tex]\( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex]:
[tex]\[ t = \frac{-4 \pm \sqrt{4^2 - 4 \cdot (-4.9) \cdot 180}}{2 \cdot (-4.9)} \][/tex]
First, calculate the discriminant (the part under the square root):
[tex]\[ \Delta = 4^2 - 4 \cdot (-4.9) \cdot 180 \][/tex]
[tex]\[ \Delta = 16 + 4 \cdot 4.9 \cdot 180 \][/tex]
[tex]\[ \Delta = 16 + 3528 \][/tex]
[tex]\[ \Delta = 3544 \][/tex]
Next, solve for [tex]\(t\)[/tex]:
[tex]\[ t = \frac{-4 \pm \sqrt{3544}}{-9.8} \][/tex]
[tex]\[ \sqrt{3544} \approx 59.56 \][/tex]
Thus, we have two solutions for [tex]\(t\)[/tex]:
[tex]\[ t_1 = \frac{-4 + 59.56}{-9.8} \][/tex]
[tex]\[ t_1 \approx \frac{55.56}{-9.8} \][/tex]
[tex]\[ t_1 \approx -5.67 \, \text{seconds} \][/tex]
[tex]\[ t_2 = \frac{-4 - 59.56}{-9.8} \][/tex]
[tex]\[ t_2 \approx \frac{-63.56}{-9.8} \][/tex]
[tex]\[ t_2 \approx 6.48 \, \text{seconds} \][/tex]
Since time cannot be negative, we discard [tex]\(t_1\approx -5.67\)[/tex] and consider [tex]\(t_2 \approx 6.48\)[/tex].
We'll now round [tex]\(6.48\)[/tex] to the nearest given option. The options are: 5, 6.5, 8.5, 9, 10.
The closest option to [tex]\(6.48\)[/tex] is [tex]\(6.5\)[/tex].
Therefore, the rocket strikes the ground after approximately:
(B) 6.5 seconds
