Let's analyze the problem based on the image:

The height of the rocket as a function of time $t$ (in seconds) is given by the quadratic equation:

$h(t) = -16t^2 + 160t + 176$

Part a: Initial Height

To find the initial height of the rocket, we substitute $t = 0$ into the height equation:

$h(0) = -16(0)^2 + 160(0) + 176 = 176 \, \text{feet}$

Thus, the rocket's initial height is 176 feet.

Part b: Time to Reach the Ground

To find when the rocket reaches the ground, we need to solve for $t$ when $h(t) = 0$:

$-16t^2 + 160t + 176 = 0$

This is a quadratic equation. We can use the quadratic formula:

$t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

where $a = -16$, $b = 160$, and $c = 176$.

Substitute the values into the quadratic formula:

$t = \frac{-160 \pm \sqrt{160^2 - 4(-16)(176)}}{2(-16)}$ $t = \frac{-160 \pm \sqrt{25600 + 11264}}{-32}$ $t = \frac{-160 \pm \sqrt{36864}}{-32}$ $t = \frac{-160 \pm 192}{-32}$

Now we have two solutions: $t_1 = \frac{-160 + 192}{-32} = \frac{32}{-32} = -1 \, \text{(ignore negative time)}$ $t_2 = \frac{-160 - 192}{-32} = \frac{-352}{-32} = 11$

Thus, the rocket reaches the ground after 11 seconds.

Final Answers:

• a. The rocket's initial height is 176 feet.
• b. The rocket reaches the ground at t = 11 seconds.

Do you have any questions or want more details?

Follow-up questions:

1. How would the height equation change if the rocket were launched with a different initial velocity?
2. What does the coefficient $-16$ represent in this equation?
3. What is the maximum height of the rocket and when does it occur?
4. How would you find the time at which the rocket reaches its maximum height?
5. What would happen if the initial height was zero?

Tip:

Always check the signs of your coefficients when applying the quadratic formula—negative coefficients can affect the roots!