Let's tackle these problems step by step! Here's the breakdown and solution for each question:

5. Domain of the function $h(t) = -16t^2 + 80t$:

• The height function models the projectile's motion.
• To find the domain, determine the time interval $t$ during which the object is in motion (when $h(t) \geq 0$).

1. Solve $h(t) = -16t^2 + 80t = 0$: Factorize: $t(-16t + 80) = 0 \implies t = 0 \, \text{or} \, t = 5$ The object is on the ground at $t = 0$ and $t = 5$.

2. Since $h(t)$ represents height, the domain is: $[0, 5] \, \text{seconds}.$

6. Maximum height for $h(t) = -16t^2 + 110t + 10$:

• This quadratic equation is downward-opening ($-16t^2$), so the maximum height occurs at the vertex.

1. Find the time of the vertex: $t = -\frac{b}{2a} = -\frac{110}{2(-16)} = \frac{110}{32} = 3.4375 \, \text{seconds}.$

2. Substitute $t = 3.4375$ into $h(t)$: $h(3.4375) = -16(3.4375)^2 + 110(3.4375) + 10 = 198.06 \, \text{feet}.$ The maximum height is approximately 198.06 feet.

7. Eva's trampoline jump $h(t) = -16t^2 + 20t + 6$:

• Determine if Eva reaches $h(t) = 14$ feet.

1. Solve $-16t^2 + 20t + 6 = 14$: Simplify: $-16t^2 + 20t - 8 = 0 \implies -8(2t^2 - 2.5t + 1) = 0.$ Divide through by $-8$: $2t^2 - 2.5t + 1 = 0.$

2. Use the quadratic formula: $t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}, \, a = 2, b = -2.5, c = 1.$ $t = \frac{-(-2.5) \pm \sqrt{(-2.5)^2 - 4(2)(1)}}{2(2)} = \frac{2.5 \pm \sqrt{6.25 - 8}}{4}.$ $t = \frac{2.5 \pm \sqrt{-1.75}}{4}.$ Since the discriminant ($-1.75$) is negative, **Eva does not reach