Let's analyze the given problem.

The height of the rocket above sea level is given by the quadratic function:

$h(t) = -4.9t^2 + 193t + 157$

where:

• $h(t)$ is the height in meters,
• $t$ is the time in seconds since launch.

Questions to Solve:

1. Splashdown Time: Find when the rocket hits the ocean ($h(t) = 0$).
2. Maximum Height: Find the peak height of the rocket ($h(t)$ at its maximum).
3. Launchpad Height: Find the height at $t = 0$.

Let me calculate these step by step.

1. When does the rocket splash down ($h(t) = 0$)?

Set $h(t) = 0$: $-4.9t^2 + 193t + 157 = 0$

Solve this quadratic equation using the quadratic formula: $t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ Here:

• $a = -4.9$,
• $b = 193$,
• $c = 157$.

Substitute these values: $t = \frac{-193 \pm \sqrt{193^2 - 4(-4.9)(157)}}{2(-4.9)}$

Compute $\Delta = b^2 - 4ac$: $\Delta = 193^2 - 4(-4.9)(157) = 37249 + 3077.2 = 40326.2$

Substitute $\Delta$ into the formula: $t = \frac{-193 \pm \sqrt{40326.2}}{-9.8}$

Approximate: $\sqrt{40326.2} \approx 200.82$ $t = \frac{-193 \pm 200.82}{-9.8}$

Solve for both roots: $t_1 = \frac{-193 + 200.82}{-9.8} = \frac{7.82}{-9.8} \approx -0.798 \, \text{(not valid for time)}$ $t_2 = \frac{-193 - 200.82}{-9.8} = \frac{-393.82}{-9.8} \approx 40.186$

Thus, the rocket splashes down after approximately: $t \approx 40.186 \, \text{seconds}.$

2. Maximum Height of the Rocket

The maximum height occurs at the vertex of the parabola. The formula for the time of the vertex is: $t = -\frac{b}{2a}$

Substitute $a = -4.9$ and $b = 193$: $t = -\frac{193}{2(-4.9)} = \frac{193}{9.8} \approx 19.694 \, \text{seconds}.$

Substitute $t = 19.694$ into $h(t)$ to find the height: $h(19.694) = -4.9(19.694)^2 + 193(19.694) + 157$

Calculate step by step: $19.694^2 \approx 387.06$ $h(19.694) = -4.9(387.06) + 193(19.694) + 157$ $h(19.694) = -1896.594 + 3793.942 + 157 \approx 2054.348 \, \text{meters}.$

Thus, the rocket reaches a maximum height of approximately: $h \approx 2054.348 \, \text{meters}.$

3. Launchpad Height

The height at $t = 0$ is: $h(0) = -4.9(0)^2 + 193(0) + 157 = 157 \, \text{meters}.$

Thus, the rocket is launched from a height of: $h = 157 \, \text{meters}.$

Final Answers:

1. Splashdown Time: $t \approx 40.186 \, \text{seconds}$.
2. Maximum Height: $h \approx 2054.348 \, \text{meters}$.
3. Launchpad Height: $h = 157 \, \text{meters}$.

Do you need further clarification on any of these steps, or would you like a more detailed breakdown?

Here are 5 related questions you could explore:

1. What is the velocity of the rocket when it reaches the ocean?
2. Can we calculate how long it takes to reach the maximum height?
3. How does changing the initial velocity affect the maximum height?
4. What if air resistance were included? How would it alter the results?
5. Can you derive the total distance traveled by the rocket?

Tip: Always check the discriminant ($\Delta$) in quadratic equations to ensure the roots are real and valid!