The height (in meters) of a baseball follows a trajectory given by $h(t) = -4.9t^2 + 14t - 0.4$ at time $t$ (in seconds). As an improper fraction, for how long is the baseball above a height of $8$ meters?
To find when the baseball is above a height of $8$ meters, we must solve the inequality $h(t) > 8$. Substituting the expression for $h(t)$ into the inequality gives $-4.9t^2 + 14t - 0.4 > 8$. Rewriting this inequality by moving all the terms to the left-hand side gives $-4.9t^2 + 14t - 8.4 > 0$. This inequality is made easier to work with by multiplying everything by $10$ to clear decimals. Doing so yields $-49t^2 +140t -84 > 0$.
This quadratic takes a downward-opening parabolic shape (because the leading coefficient, $-49$, is negative) and will therefore be positive outside the interval(s) it is equal to $0$. To find when it is equal to $0$, we split the quadratic into its linear binomial factors by setting it equal to $0$ and applying factoring techniques. We look for two numbers that multiply to give $-49(-84) = 4116$ and add to give $140$. After some guess and check, it appears that $69 \cdot 60 = 4140$ is close, so we write $\pm 69 \cdot \pm 60$. Factoring, $$-49t^2 + 140t - 84 = (-49t + 69)(t - 60).$$Setting each factor equal to $0$ yields valid roots $t = \frac{69}{49} = \frac{3}{7}$ and $t = 60$.
The quadratic has roots $\frac{3}{7}$ and $60$, so it changes sign at each endpoint and between the two roots. From this, we can quickly deduce that it is positive when $t < \frac{3}{7}$ or $t > 60$ and negative when $\frac{3}{7} < t < 60$. Thus, the inequality $-4.9t^2 + 14t - 8.4 > 0$ is satisfied when $\boxed{\frac{3}{7} < t < 60}$.
To find for how long the baseball is above a height of 8 meters, we need to solve the equation $h(t) > 8$ for $t$.
Replacing $h(t)$ with the given function, we have:
$-4.9t^2 + 14t - 0.4 > 8$
Let's rearrange the equation to isolate the quadratic term:
$-4.9t^2 + 14t - 0.4 - 8 > 0$
Combine like terms:
$-4.9t^2 + 14t - 8.4 > 0$
Now, to solve this inequality, we can use the quadratic formula. The quadratic formula states that for a quadratic equation of the form $ax^2 + bx + c = 0$, the solutions for $x$ are given by:
$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
In our case, the quadratic equation is $-4.9t^2 + 14t - 8.4 = 0$, so we have:
$t = \frac{-14 \pm \sqrt{14^2 - 4(-4.9)(-8.4)}}{2(-4.9)}$
Calculating the values inside the square root:
$t = \frac{-14 \pm \sqrt{196 - (4)(-4.9)(-8.4)}}{2(-4.9)}$
Simplifying further:
$t = \frac{-14 \pm \sqrt{196 - 163.68}}{2(-4.9)}$
$t = \frac{-14 \pm \sqrt{196 + 163.68}}{2(-4.9)}$
$t = \frac{-14 \pm \sqrt{359.68}}{2(-4.9)}$
$t = \frac{-14 \pm \sqrt{359.68}}{-9.8}$
Now, let's evaluate the square root:
$t = \frac{-14 \pm 18.97}{-9.8}$
The two possible values for $t$ are:
$t_1 = \frac{-14 + 18.97}{-9.8}$
$t_1 = \frac{4.97}{-9.8}$
$t_2 = \frac{-14 - 18.97}{-9.8}$
$t_2 = \frac{-32.97}{-9.8}$
Simplifying each fraction:
$t_1 = -0.507$
$t_2 = 3.362$
As the question asks for the time the baseball is above 8 meters, we need to take into consideration only the positive values of time.
Therefore, for $t > 0$, the baseball is above a height of 8 meters for approximately $3.362$ seconds.