To determine how long the ball is at or above a height of 38 feet, we start with the height equation given:

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

We need to find the times $t$ when $h(t) \geq 38$. Setting up the inequality:

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

Subtracting 38 from both sides gives:

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

Next, we simplify this equation:

$-16t^2 + 48t - 32 = 0$

Dividing through by -16 (and reversing the inequality):

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

Now we can factor this quadratic:

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

To find the critical points, we set each factor to zero:

$t - 1 = 0 \quad \Rightarrow \quad t = 1$ $t - 2 = 0 \quad \Rightarrow \quad t = 2$

The quadratic will be zero at $t = 1$ and $t = 2$. We can analyze the intervals around these points to find where the expression is less than or equal to zero.

1. For $t < 1$: Choose $t = 0$: $(0 - 1)(0 - 2) = 2 > 0$

2. For $1 < t < 2$: Choose $t = 1.5$: $(1.5 - 1)(1.5 - 2) = (0.5)(-0.5) = -0.25 < 0$

3. For $t > 2$: Choose $t = 3$: $(3 - 1)(3 - 2) = (2)(1) = 2 > 0$

The quadratic is negative or zero in the interval:

$1 \leq t \leq 2$

Thus, the ball is at or above a height of 38 feet for:

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

Summary

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

Would you like any further details or clarifications? Here are some related questions you might consider:

1. What is the maximum height reached by the ball?
2. How long does it take for the ball to reach its maximum height?
3. What is the time when the ball first hits the ground?
4. How does the launch speed affect the time the ball is above a certain height?
5. What would be the height of the ball after 3 seconds?

Tip: When analyzing quadratic functions, consider the vertex and intercepts to understand the shape and behavior of the graph.