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Let's solve each part of the question step by step, given the height function:

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

where \( h \) is the height of the ball above the ground (in meters) at time \( t \) seconds.

### Part (a): Find the time (in seconds) when the ball reaches its maximum height.

To find the time at which the ball reaches its maximum height, we need to calculate the vertex of the quadratic function. Since the function is in the form \( h = at^2 + bt + c \), the time at the vertex (maximum height in this case because \( a < 0 \)) is given by:

\[
t = -\frac{b}{2a}
\]

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

Substituting these values:

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

So, the ball reaches its maximum height at approximately **0.94 seconds**.

### Part (b): Find the maximum height of the ball.

Now that we know the time at which the maximum height occurs, we can substitute \( t = 0.9375 \) back into the height equation to find the maximum height.

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

Calculating this:

1. \( 9 \times 0.9375 = 8.4375 \)
2. \( 4.8 \times (0.9375)^2 = 4.21875 \)

So,

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

Thus, the maximum height of the ball is approximately **6.22 meters**.

### Part (c): After how much time does the ball hit the ground?

The ball hits the ground when \( h = 0 \). Therefore, we set the equation to zero and solve for \( t \):

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

Rearranging, we get:

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

This is a quadratic equation in the form \( at^2 + bt + c = 0 \). We can solve it using the quadratic formula:

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

where:
- \( a = 4.8 \)
- \( b = -9 \)
- \( c = -2 \)

Substituting these values:

\[
t = \frac{-(-9) \pm \sqrt{(-9)^2 - 4 \cdot 4.8 \cdot (-2)}}{2 \cdot 4.8}
\]

\[
t = \frac{9 \pm \sqrt{81 + 38.4}}{9.6}
\]

\[
t = \frac{9 \pm \sqrt{119.4}}{9.6}
\]

Calculating further:

1. \( \sqrt{119.4} \approx 10.927 \)
2. \( t = \frac{9 \pm 10.927}{9.6} \)

This gives two solutions for \( t \):

- \( t = \frac{9 + 10.927}{9.6} \approx 2.08 \) seconds
- \( t = \frac{9 - 10.927}{9.6} \approx -0.20 \) seconds (which we discard as time cannot be negative)

So, the ball hits the ground approximately **2.08 seconds** after it was thrown.

---

Would you like a more detailed breakdown of any part, or do you have any specific questions?

Here are five related questions for further practice:

1. What would be the height of the ball at \( t = 1 \) second?
2. If the initial velocity of the throw were different, how would that affect the maximum height?
3. How would the time to reach the maximum height change if gravity were stronger?
4. How could you adjust the equation to reflect a throw from a height of 3 meters instead?
5. How would you find the speed of the ball at its maximum height?

**Tip:** The derivative of the height function gives the velocity function, which helps find when the object is momentarily stationary (at maximum height) or moving in different directions.

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?

Learn how to find the maximum height and impact time of a ball thrown vertically, modeled by the quadratic function h = 2 + 9t - 4.8t^2. This step-by-step solution includes vertex and quadratic formulas, suitable for Grade 10 students.

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