From the provided image, let’s analyze and solve the problem step by step for clarity:

Given:

The function for Jason's height is: $h(t) = -16t^2 + 16t + 480$ where $t$ is in seconds, and $h(t)$ is in feet.

(a) How long did it take for Jason to reach his maximum height?

The time to reach the maximum height is found by using the formula for the vertex of a parabola: $t = \frac{-b}{2a}$ Here:

• $a = -16$
• $b = 16$

Substituting: $t = \frac{-16}{2(-16)} = \frac{-16}{-32} = 0.5 \, \text{seconds}.$

Answer: $0.5 \, \text{seconds}$

(b) What was the highest point that Jason reached?

To find the maximum height, substitute $t = 0.5$ into $h(t)$: $h(0.5) = -16(0.5)^2 + 16(0.5) + 480$

Calculate each term:

• $(0.5)^2 = 0.25$
• $-16(0.25) = -4$
• $16(0.5) = 8$

Adding these terms: $h(0.5) = -4 + 8 + 480 = 484 \, \text{feet}.$

Answer: $484 \, \text{feet}$

(c) Jason hit the water after how many seconds?

To find when Jason hits the water, solve for $t$ when $h(t) = 0$: $-16t^2 + 16t + 480 = 0$

Simplify by dividing through by $-16$: $t^2 - t - 30 = 0$

Factorize: $(t - 6)(t + 5) = 0$

Solutions: $t = 6 \quad \text{or} \quad t = -5$

Since time cannot be negative: $t = 6 \, \text{seconds}.$

Answer: $6 \, \text{seconds}$

Summary of Answers:

(a) $0.5 \, \text{seconds}$
(b) $484 \, \text{feet}$
(c) $6 \, \text{seconds}$

Do you want more details about any part of the solution, or have questions? Here are 5 additional questions you could consider:

1. How would the equation change if Jason jumped from a higher platform?
2. How can you derive the quadratic formula for solving part (c)?
3. What does the negative coefficient of $t^2$ tell you about the graph of $h(t)$?
4. Can you compute the time at which Jason was halfway down to the water?
5. How would air resistance affect the height function?

Tip: Always double-check your factorizations or vertex calculations for quadratic equations!