Question: The equation $y = -16t^2 + 32t + 5$ represents the height of a projectile, $y$, in feet at a particular time $t$, in seconds.
For what interval (or intervals) of time will the projectile’s height be below 20 feet?
1. **State the problem:** We want to find the time intervals when the height $y$ of the projectile is less than 20 feet.
2. **Write the inequality:**
$$-16t^2 + 32t + 5 < 20$$
3. **Bring all terms to one side:**
$$-16t^2 + 32t + 5 - 20 < 0$$
$$-16t^2 + 32t - 15 < 0$$
4. **Solve the quadratic inequality:** First, solve the equation
$$-16t^2 + 32t - 15 = 0$$
5. **Use the quadratic formula:**
$$t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$
where $a = -16$, $b = 32$, and $c = -15$.
Calculate the discriminant:
$$\Delta = 32^2 - 4(-16)(-15) = 1024 - 960 = 64$$
Calculate the roots:
$$t = \frac{-32 \pm \sqrt{64}}{2(-16)} = \frac{-32 \pm 8}{-32}$$
6. **Find each root:**
$$t_1 = \frac{-32 + 8}{-32} = \frac{-24}{-32} = \frac{3}{4} = 0.75$$
$$t_2 = \frac{-32 - 8}{-32} = \frac{-40}{-32} = \frac{5}{4} = 1.25$$
7. **Analyze the inequality:** Since $a = -16 < 0$, the parabola opens downward.
The quadratic expression is less than zero outside the roots:
$$t < 0.75 \quad \text{or} \quad t > 1.25$$
8. **Interpretation:** The projectile's height is below 20 feet before $0.75$ seconds and after $1.25$ seconds.
**Final answer:** The height is below 20 feet for
$$t < 0.75 \quad \text{and} \quad t > 1.25$$