Let’s work through the parts of this problem step-by-step.

---

Part 1: Function for height $s(t)$

The motion of the rocket is governed by the formula for height in projectile motion:
$s(t) = -16t^2 + v_0t + h_0$
where:

• $-16t^2$ is the acceleration due to gravity (in feet/second²),
• $v_0 = 199$ ft/s is the initial velocity,
• $h_0 = 141$ ft is the initial height.

Thus, the function describing the height is: $s(t) = -16t^2 + 199t + 141$

---

Part 2: Time of maximum height and maximum height

The time $t$ at which the rocket reaches its maximum height occurs at the vertex of the parabola, which is calculated as: $t = -\frac{b}{2a}$ Here, $a = -16$ and $b = 199$. Substituting: $t = -\frac{199}{2(-16)} = \frac{199}{32} \approx 6.22 \, \text{seconds.}$

To find the maximum height, substitute $t = 6.22$ into $s(t)$: $s(6.22) = -16(6.22)^2 + 199(6.22) + 141$ $s(6.22) \approx -16(38.6884) + 199(6.22) + 141$ $s(6.22) \approx -619.01 + 1237.78 + 141 \approx 761.77 \, \text{feet.}$

Thus:

• Maximum height: 761.77 feet
• Time to reach max height: 6.22 seconds

---

Part 3: Time interval when the rocket is above 733 feet

To find when $s(t) > 733$, solve: $-16t^2 + 199t + 141 = 733$ Simplify: $-16t^2 + 199t - 592 = 0$ Solve this quadratic equation using the quadratic formula: $t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ Here, $a = -16$, $b = 199$, $c = -592$. Substituting: $t = \frac{-199 \pm \sqrt{199^2 - 4(-16)(-592)}}{2(-16)}$ $t = \frac{-199 \pm \sqrt{39601 - 37952}}{-32}$ $t = \frac{-199 \pm \sqrt{1649}}{-32}$ $t = \frac{-199 \pm 40.6}{-32}$ $t_1 = \frac{-199 + 40.6}{-32} \quad \text{and} \quad t_2 = \frac{-199 - 40.6}{-32}$ $t_1 \approx 4.95 \, \text{seconds,} \quad t_2 \approx 7.49 \, \text{seconds.}$

The rocket is above 733 feet between 4.95 seconds and 7.49 seconds.

---

Part 4: Time to hit the ground

The rocket hits the ground when $s(t) = 0$: $-16t^2 + 199t + 141 = 0$ Using the quadratic formula: $t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ Substitute $a = -16$, $b = 199$, $c = 141$: $t = \frac{-199 \pm \sqrt{199^2 - 4(-16)(141)}}{2(-16)}$ $t = \frac{-199 \pm \sqrt{39601 + 9024}}{-32}$ $t = \frac{-199 \pm \sqrt{48625}}{-32}$ $t = \frac{-199 \pm 220.5}{-32}$ $t_1 = \frac{-199 + 220.5}{-32} \quad \text{and} \quad t_2 = \frac{-199 - 220.5}{-32}$ $t_1 \approx -0.67 \, \text{(not valid)} \quad \text{and} \quad t_2 \approx 13.11 \, \text{seconds.}$

The rocket hits the ground after approximately 13.11 seconds.

---

Final Answers

a) Function: $s(t) = -16t^2 + 199t + 141$
b) Maximum height: $761.77$ feet at $t = 6.22$ seconds
c) Above 733 feet between $t = 4.95$ and $t = 7.49$ seconds
d) Hits the ground at $t = 13.11$ seconds

---

Would you like further details or assistance?
Here are 5 related questions to explore:

1. How is the quadratic formula derived, and how does it apply to this problem?
2. What is the significance of gravity ($-16t^2$) in the function?
3. How would the problem change with a different initial velocity?
4. What if the building height were doubled—how would it affect the results?
5. Can you graph $s(t)$ to visualize the rocket's motion?

Tip: When solving quadratic equations, always check if factoring or completing the square is simpler than using the quadratic formula.