Answer:
6.11 seconds
Step-by-step explanation:
Given
[tex]f(t) = -4.9t^2 + 28t + 12[/tex]
Required
When will it hit the ground?
When it hits the ground, f(t) = 0.
So, we have:
[tex]0 = -4.9t^2 + 28t + 12[/tex]
Solve quadratic equation using formula
[tex]t = \frac{-b \± \sqrt{b^2 - 4ac}}{2a}[/tex]
[tex]t = \frac{-28 \± \sqrt{28^2 - 4*-4.9*12}}{2*-4.9}[/tex]
[tex]t = \frac{-28 \± \sqrt{1019.2}}{-9.8}[/tex]
[tex]t = \frac{-28 \± 31.92}{-9.8}[/tex]
Split
[tex]t = \frac{-28 + 31.92}{-9.8}[/tex] or [tex]t = \frac{-28 - 31.92}{-9.8}[/tex]
[tex]t = \frac{3.92}{-9.8}[/tex] or [tex]t = \frac{-59.92}{-9.8}[/tex]
[tex]t = -0.4[/tex] or [tex]t = 6.11[/tex]
Time can not be negative. So:
[tex]t = 6.11[/tex]