Let's analyze the question provided in the image. The height $h$ of the ball as a function of time $t$ is given by:

$h = 30t - 5t^2$

We will solve the two parts:

(a) How long is the ball in the air?

To determine the total time the ball is in the air, we solve for $t$ when $h = 0$:

$30t - 5t^2 = 0$

Factorize:

$t(30 - 5t) = 0$

This gives two solutions:

$t = 0 \quad \text{or} \quad t = \frac{30}{5} = 6$

The ball is in the air for 6 seconds.

(b) Find the greatest height the ball reaches.

To find the maximum height, we determine the vertex of the parabola described by the equation $h = 30t - 5t^2$. The time at which the maximum height occurs is given by:

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

Here, $a = -5$ and $b = 30$:

$t = -\frac{30}{2(-5)} = 3 \, \text{seconds}$

Substitute $t = 3$ into the height equation to find the maximum height:

$h = 30(3) - 5(3)^2$

$h = 90 - 45 = 45 \, \text{meters}$

The greatest height the ball reaches is 45 meters.

Final Answers:

(a) The ball is in the air for 6 seconds.
(b) The greatest height the ball reaches is 45 meters.

Would you like additional explanations or graphs? Here are some follow-up questions:

1. How can you graph the equation $h = 30t - 5t^2$?
2. What does the negative coefficient of $t^2$ signify about the ball's motion?
3. How would air resistance alter the equations of motion?
4. Can you determine the velocity of the ball at any time $t$?
5. How is the vertex formula related to real-world problems like this one?

Tip: Always confirm the roots of the equation when solving quadratic problems to ensure both solutions make sense in the context of the problem.