To solve this problem, we need to find the maximum height of the ball based on the height equation provided. The equation of the height $h(t)$ of the ball as a function of time $t$ is:

$h(t) = 3 + 12t - 2t^2$

Step 1: Find the first derivative

The first step is to find the critical points by taking the derivative of the height function and setting it equal to zero. This will help us identify the time $t$ at which the maximum height occurs.

$\frac{dh}{dt} = 12 - 4t$

Step 2: Set the derivative equal to zero

To find the critical points, we set the derivative equal to zero:

$12 - 4t = 0$

Solving for $t$:

$4t = 12$ $t = 3 \, \text{seconds}$

Step 3: Verify if it's a maximum

Now, take the second derivative to check if this is a maximum point:

$\frac{d^2h}{dt^2} = -4$

Since the second derivative is negative, this confirms that $t = 3$ seconds corresponds to a maximum height.

Step 4: Find the maximum height

Substitute $t = 3$ back into the original height function:

$h(3) = 3 + 12(3) - 2(3^2)$ $h(3) = 3 + 36 - 18 = 21 \, \text{metres}$

Thus, the maximum height of the ball is 21 metres.

Would you like more details or have any questions?

Here are 5 questions to deepen understanding:

1. How do we confirm whether a critical point is a maximum or minimum?
2. Why do we set the first derivative equal to zero to find the maximum height?
3. What does the second derivative tell us about the shape of the function?
4. How would the maximum height change if the equation were different?
5. Can we use this method for other types of motion, such as projectile motion?

Tip: To find maximum or minimum points in any function, always look at the first and second derivatives!