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The question provided in the image is:

A ball is thrown from an initial height of 2 feet with an initial upward velocity of \(34 \, \text{ft/s}\). The ball's height \(h\) (in feet) after \(t\) seconds is given by the equation:

\[
h = 2 + 34t - 16t^2
\]

Find all values of \(t\) for which the ball's height is \(18\) feet. Round answers to the nearest hundredth if needed.

---

### Step 1: Set up the equation
We are solving for \(t\) when \(h = 18\):
\[
18 = 2 + 34t - 16t^2
\]

### Step 2: Rearrange into standard quadratic form
Subtract \(18\) from both sides:
\[
0 = 2 - 18 + 34t - 16t^2
\]
\[
0 = -16t^2 + 34t - 16
\]
Rewriting:
\[
-16t^2 + 34t - 16 = 0
\]

### Step 3: Solve the quadratic equation
Divide through by \(-2\) to simplify:
\[
8t^2 - 17t + 8 = 0
\]

We solve this using the quadratic formula:
\[
t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
Here:
- \(a = 8\)
- \(b = -17\)
- \(c = 8\)

Substitute these values:
\[
t = \frac{-(-17) \pm \sqrt{(-17)^2 - 4(8)(8)}}{2(8)}
\]
\[
t = \frac{17 \pm \sqrt{289 - 256}}{16}
\]
\[
t = \frac{17 \pm \sqrt{33}}{16}
\]

### Step 4: Calculate the roots
Approximate \(\sqrt{33} \approx 5.744\):
\[
t = \frac{17 \pm 5.744}{16}
\]

#### First root:
\[
t = \frac{17 + 5.744}{16} = \frac{22.744}{16} \approx 1.42
\]

#### Second root:
\[
t = \frac{17 - 5.744}{16} = \frac{11.256}{16} \approx 0.70
\]

---

### Final Answer:
The ball's height is \(18\) feet at:
\[
t \approx 0.70 \, \text{seconds and } t \approx 1.42 \, \text{seconds.}
\]

Would you like a breakdown of the calculation or additional related questions? 

---

### Related Questions:
1. How can the quadratic formula be applied in other physics-related problems?
2. What does the discriminant \((b^2 - 4ac)\) tell us about the motion of the ball?
3. How would the graph of \(h = 2 + 34t - 16t^2\) look, and what does it signify?
4. What happens to the ball after \(t = 1.42 \, \text{seconds}\)?
5. How can you determine the maximum height of the ball using calculus?

---

### Tip:
When solving quadratic equations, always double-check your discriminant calculation to avoid errors in finding the roots.

A ball is thrown from an initial height of 2 feet with an initial upward velocity of 34 ft/s. The ball's height h (in feet) after t seconds is given by the equation: h = 2 + 34t - 16t^2. Find all values of t for which the ball's height is 18 feet. Round answers to the nearest hundredth if needed.

Learn how to solve a quadratic equation to determine the time when a ball reaches a height of 18 feet. The problem involves algebra, quadratic equations, and projectile motion concepts. Suitable for high school students in Grades 9-12.

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