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?

Answer 1

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 $\boxed{\frac{3}{7} < t < 60}$.