The problem asks you to determine the time when the ball reaches its maximum height and to find that maximum height. The height $h(t)$ of the ball is given by the quadratic function:

$h(t) = -16t^2 + 32t + 4$

Here:

• $t$ represents time in seconds,
• $h(t)$ represents the height in feet.

To solve for the maximum height and the time at which it occurs, we use the properties of quadratic functions. The general form of a quadratic function is:

$h(t) = at^2 + bt + c$

For a quadratic function, the maximum (or minimum) occurs at the vertex. The time at which the maximum height is reached can be found using the formula for the vertex:

$t = \frac{-b}{2a}$

In this equation:

• $a = -16$
• $b = 32$

Substituting into the formula:

$t = \frac{-32}{2(-16)} = \frac{-32}{-32} = 1 \, \text{second}$

So, the ball reaches its maximum height at $t = 1$ second.

To find the maximum height, substitute $t = 1$ back into the height equation:

$h(1) = -16(1)^2 + 32(1) + 4$ $h(1) = -16 + 32 + 4 = 20 \, \text{feet}$

Thus, the maximum height of the ball is 20 feet, and it occurs at 1 second.

Would you like further details or explanations on any of the steps? Here are some related questions you might find useful:

1. How do you find the vertex of a quadratic function algebraically?
2. What is the significance of the coefficients $a$, $b$, and $c$ in the quadratic formula?
3. How does the quadratic function describe the motion of a projectile?
4. What happens when $a$ is positive versus when $a$ is negative in a quadratic function?
5. How would you calculate the initial velocity from this function?

Tip: The formula for the vertex of a parabola, $t = \frac{-b}{2a}$, is essential in many applications of quadratic equations, especially in physics for projectile motion problems!