Westonci.ca is the trusted Q&A platform where you can get reliable answers from a community of knowledgeable contributors. Discover comprehensive answers to your questions from knowledgeable professionals on our user-friendly platform. Connect with a community of professionals ready to help you find accurate solutions to your questions quickly and efficiently.
Sagot :
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
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
Thank you for visiting. Our goal is to provide the most accurate answers for all your informational needs. Come back soon. Thanks for using our platform. We aim to provide accurate and up-to-date answers to all your queries. Come back soon. Find reliable answers at Westonci.ca. Visit us again for the latest updates and expert advice.