To find the time the ball is at or above a height of 38 feet, we start with the equation for the height of the ball:

$h = -16t^2 + 48t + 6$

We need to set this equation greater than or equal to 38:

$-16t^2 + 48t + 6 \geq 38$

Rearranging gives:

$-16t^2 + 48t + 6 - 38 \geq 0$

$-16t^2 + 48t - 32 \geq 0$

Dividing the entire inequality by -16 (remembering to flip the inequality sign):

$t^2 - 3t + 2 \leq 0$

Next, we factor the quadratic:

$(t - 1)(t - 2) \leq 0$

Now, we determine the intervals where this inequality holds. The roots are $t = 1$ and $t = 2$. We test the intervals around these points:

1. Interval $(-\infty, 1)$: Choose $t = 0$: $(0 - 1)(0 - 2) = 2 > 0$

2. Interval $(1, 2)$: Choose $t = 1.5$: $(1.5 - 1)(1.5 - 2) = (-0.25) < 0$

3. Interval $(2, \infty)$: Choose $t = 3$: $(3 - 1)(3 - 2) = 2 > 0$

The inequality $(t - 1)(t - 2) \leq 0$ holds for:

$1 \leq t \leq 2$

Thus, the ball is at or above a height of 38 feet from $t = 1$ second to $t = 2$ seconds.

Duration

The duration for which the ball is at or above 38 feet is:

$2 - 1 = 1 \text{ second}$

Summary

The ball is at or above a height of 38 feet for 1 second.

Would you like further details or explanations? Here are some related questions:

1. What are the maximum height and time at which it occurs?
2. How does the ball's height change after 2 seconds?
3. What would happen if the launch speed were increased?
4. How would you graph the height function?
5. What is the total time the ball is in the air?

Tip: Always check the endpoints of your intervals when solving inequalities.