Answer:
1.18 seconds
Step-by-step explanation:
To find out when the egg is 8 feet from the ground, we need to solve the equation [tex]d = -16t^2 - 4t + 35[/tex] for [tex]t[/tex] when [tex]d = 8[/tex].
So, we have:
[tex]8 = -16t^2 - 4t + 35[/tex]
Subtract 8 from both sides:
[tex]8-8 = -16t^2 - 4t + 35-8[/tex]
[tex]-16t^2 - 4t + 35 - 8 = 0[/tex]
Simplify:
[tex]-16t^2 - 4t + 27 = 0[/tex]
To solve this quadratic equation, we can use the quadratic formula:
[tex] \Large\boxed{\boxed{t = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}}}[/tex]
While comparing above equation with ax²+bx+c = 0, we get
[tex]a = -16[/tex], [tex]b = -4[/tex], and [tex]c = 27[/tex].
Substitute these values, we get:
[tex]t = \dfrac{-(-4) \pm \sqrt{(-4)^2 - 4(-16)(27)}}{2(-16)}[/tex]
[tex]t = \dfrac{4 \pm \sqrt{16 + 1728}}{-32}[/tex]
[tex]t = \dfrac{4 \pm \sqrt{1744}}{-32}[/tex]
Now, calculate the value under the square root:
[tex]\sqrt{1744} \approx 41.79[/tex]
So, we have:
[tex]t = \dfrac{4 \pm 41.79}{-32}[/tex]
Now, we have two potential solutions:
[tex]t_1 = \dfrac{4 + 41.79}{-32} \\\\ \approx −1.4309375 \\\\ \approx -1.43 \textsf{(in nearest hundredth)}[/tex]
[tex]t_2 = \dfrac{4 - 41.79}{-32} \\\\ \approx 1.1809375\\\\ \approx 1.18 \textsf{(in nearest hundredth)}[/tex]
Since time cannot be negative in this context, we discard [tex]t_1[/tex].
Thus, the egg will be 8 feet from the ground approximately 1.18 seconds after it is thrown.
