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Let's solve each part of this problem step by step, given that the height of the ball is modeled by the equation:

\[
h = 2 + 9t - 4.8t^2
\]

### Part (a): Time to Reach Maximum Height
To find the time when the ball reaches its maximum height, we need to determine the vertex of this quadratic equation. For a function in the form \( h(t) = at^2 + bt + c \), the time at which the maximum height occurs is given by \( t = -\frac{b}{2a} \).

In this equation:
- \( a = -4.8 \)
- \( b = 9 \)

Thus,

\[
t = -\frac{9}{2 \times -4.8} = \frac{9}{9.6} = 0.9375 \text{ seconds}
\]

So, the ball reaches its maximum height at approximately \( 0.94 \) seconds.

### Part (b): Maximum Height of the Ball
To find the maximum height, substitute \( t = 0.9375 \) back into the height equation:

\[
h = 2 + 9(0.9375) - 4.8(0.9375)^2
\]

Calculating each term:
1. \( 9 \times 0.9375 = 8.4375 \)
2. \( 4.8 \times (0.9375)^2 = 4.21875 \)

Then,

\[
h = 2 + 8.4375 - 4.21875 = 6.21875 \text{ meters}
\]

Thus, the maximum height of the ball is approximately \( 6.22 \) meters.

### Part (c): Time When the Ball Hits the Ground
The ball hits the ground when \( h = 0 \). Set the height equation to zero and solve for \( t \):

\[
0 = 2 + 9t - 4.8t^2
\]

This is a quadratic equation in the form \( at^2 + bt + c = 0 \), where:
- \( a = -4.8 \)
- \( b = 9 \)
- \( c = 2 \)

Using the quadratic formula:

\[
t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

Substitute the values:

\[
t = \frac{-9 \pm \sqrt{9^2 - 4 \times -4.8 \times 2}}{2 \times -4.8}
\]
\[
t = \frac{-9 \pm \sqrt{81 + 38.4}}{-9.6}
\]
\[
t = \frac{-9 \pm \sqrt{119.4}}{-9.6}
\]
\[
t = \frac{-9 \pm 10.926}{-9.6}
\]

This yields two solutions:

1. \( t = \frac{-9 + 10.926}{-9.6} = -0.200 \) (discard, as time cannot be negative)
2. \( t = \frac{-9 - 10.926}{-9.6} = 2.08 \)

Thus, the ball hits the ground after approximately \( 2.08 \) seconds.

---

Would you like more details or have any questions on this solution?

### Related Questions:
1. How would the height equation change if the ball were thrown with a different initial velocity?
2. What would be the maximum height if gravity were slightly different, say on another planet?
3. How can we determine the velocity of the ball at any given time?
4. What role does the initial height play in reaching the maximum height?
5. How would air resistance affect the time the ball reaches maximum height and hits the ground?

### Tip:
For any projectile motion, the maximum height is reached when the velocity becomes zero in the vertical direction. This simplifies calculations in many physics problems involving parabolic motion.

A man who is 2 metres tall throws a ball straight up in the air; the ball’s height above the ground at time t seconds is h = 2 + 9t - 4.8t^2 m.

a) Find the time (in seconds) when the ball reaches its maximum height.
b) Find the maximum height of the ball.
c) After how much time does the ball hit the ground?

Explore a projectile motion problem involving a ball thrown vertically. Learn how to find the time to reach maximum height, calculate the maximum height, and determine when the ball hits the ground using quadratic equations. Suitable for high school students in Grades 9-12.

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