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The problem involves a ball thrown vertically from a height of 7.2 meters above the ground with an initial velocity. The height of the ball \( h(t) \) above the ground after \( t \) seconds is given by the equation:

\[
h(t) = -4.9t^2 + 6t + 7.2
\]

The tasks involve determining:

(a) The height reached by the ball after 1 second.
(b) The time it takes for the ball to hit the ground.

Let's solve these two parts step by step.

### Part (a): Height after 1 second

We need to find the height of the ball after 1 second. Plugging \( t = 1 \) into the equation \( h(t) = -4.9t^2 + 6t + 7.2 \):

\[
h(1) = -4.9(1)^2 + 6(1) + 7.2 = -4.9 + 6 + 7.2 = 8.3 \text{ meters}
\]

### Part (b): Time to hit the ground

The ball will hit the ground when \( h(t) = 0 \). So we solve the quadratic equation:

\[
-4.9t^2 + 6t + 7.2 = 0
\]

This is a standard quadratic equation \( at^2 + bt + c = 0 \) where:
- \( a = -4.9 \)
- \( b = 6 \)
- \( c = 7.2 \)

We solve this using the quadratic formula:

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

Plugging in the values:

\[
t = \frac{-6 \pm \sqrt{6^2 - 4(-4.9)(7.2)}}{2(-4.9)}
\]

\[
t = \frac{-6 \pm \sqrt{36 + 141.12}}{-9.8}
\]

\[
t = \frac{-6 \pm \sqrt{177.12}}{-9.8}
\]

\[
t = \frac{-6 \pm 13.31}{-9.8}
\]

This gives us two possible times:

\[
t_1 = \frac{-6 + 13.31}{-9.8} = -0.75 \text{ seconds (not physically meaningful, as time cannot be negative)}
\]

\[
t_2 = \frac{-6 - 13.31}{-9.8} = 1.97 \text{ seconds}
\]

Thus, the ball hits the ground after approximately \( 1.97 \) seconds.

Would you like more details or have any questions about this? 

Here are some related questions you might consider:

1. What is the maximum height reached by the ball?
2. How long does it take for the ball to reach its maximum height?
3. What is the velocity of the ball at the moment it hits the ground?
4. How would the time to hit the ground change if the initial velocity were doubled?
5. What is the significance of the discriminant in the quadratic formula in this context?

**Tip:** For quadratic equations, the discriminant (\(b^2 - 4ac\)) can tell you the nature of the roots: real and distinct, real and equal, or complex.

A ball is thrown vertically upward from a height of 7.2 meters above the ground with an initial velocity of 6 m/s. The height of the ball above the ground after t seconds is given by the equation h(t) = -4.9t^2 + 6t + 7.2. Determine the height reached by the ball after 1 second, the time it takes for the ball to hit the ground, and the maximum height reached by the ball.

Explore the analysis of a ball thrown vertically upward from a height of 7.2 meters with an initial velocity of 6 m/s. Solve for the height after 1 second, the time when the ball hits the ground, and its maximum height using quadratic equations. Ideal for students in Grades 9-10.

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