To determine after how many seconds the kangaroo lands on the ground, we need to solve the equation [tex]\( d = 4 + 6t - 0.5 \cdot 9.8 \cdot t^2 \)[/tex] for [tex]\( d = 0 \)[/tex].
Firstly, let's set up the equation:
[tex]\[ 0 = 4 + 6t - 0.5 \cdot 9.8 \cdot t^2 \][/tex]
Simplifying the constants:
[tex]\[ 0 = 4 + 6t - 4.9t^2 \][/tex]
This is a quadratic equation of the form [tex]\( 0 = at^2 + bt + c \)[/tex], where [tex]\( a = -4.9 \)[/tex], [tex]\( b = 6 \)[/tex], and [tex]\( c = 4 \)[/tex].
To solve this quadratic equation, we use the quadratic formula [tex]\( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex].
Here, [tex]\( a = -4.9 \)[/tex], [tex]\( b = 6 \)[/tex], and [tex]\( c = 4 \)[/tex]. Plugging these values into the quadratic formula:
[tex]\[ t = \frac{-6 \pm \sqrt{6^2 - 4(-4.9)(4)}}{2(-4.9)} \][/tex]
[tex]\[ t = \frac{-6 \pm \sqrt{36 + 78.4}}{-9.8} \][/tex]
[tex]\[ t = \frac{-6 \pm \sqrt{114.4}}{-9.8} \][/tex]
Next, compute the square root of 114.4:
[tex]\[ \sqrt{114.4} \approx 10.694 \][/tex]
So now we have:
[tex]\[ t = \frac{-6 \pm 10.694}{-9.8} \][/tex]
This gives us two solutions:
[tex]\[ t_1 = \frac{-6 + 10.694}{-9.8} \approx \frac{4.694}{-9.8} \approx -0.479 \][/tex]
[tex]\[ t_2 = \frac{-6 - 10.694}{-9.8} \approx \frac{-16.694}{-9.8} \approx 1.704 \][/tex]
Since time cannot be negative, we discard the negative solution [tex]\( t_1 \approx -0.479 \)[/tex] seconds.
Therefore, the approximate time it takes for the kangaroo to land on the ground is 1.704 seconds.
Thus, the correct answer is:
(C) 2 seconds
