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We are given the equation that represents the height of the ball during a soccer game:

\[
h(t) = -5t^2 + 22t + 1.2
\]

Where:
- \( h(t) \) is the height of the ball in meters,
- \( t \) is the time in seconds.

We need to find the height of the ball at \( t = 3.6 \) seconds.

Let's calculate \( h(3.6) \) by substituting \( t = 3.6 \) into the equation:

\[
h(3.6) = -5(3.6)^2 + 22(3.6) + 1.2
\]

I'll calculate the value now.The height of the ball at 3.6 seconds is approximately 15.6 meters.

Would you like further details or explanations? Here are some related questions to deepen understanding:

1. What does each term in the function \( h(t) = -5t^2 + 22t + 1.2 \) represent in terms of motion?
2. How can we determine the maximum height the ball reaches during its flight?
3. At what time does the ball hit the ground?
4. How would the equation change if the initial velocity or the gravitational force were different?
5. Can we derive the velocity of the ball at any given time using this equation?

**Tip**: The term \( -5t^2 \) comes from the acceleration due to gravity, which is responsible for slowing down and eventually reversing the ball's upward motion.

During a soccer game, a player kicks the ball, and its height is expressed by the function h(t) = -5t^2 + 22t + 1.2, where h is the height in meters and t is time in seconds. What is the height of the ball at t = 3.6 seconds?

Discover how to calculate the height of a soccer ball at 3.6 seconds using the quadratic equation h(t) = -5t^2 + 22t + 1.2. Learn key concepts in algebra and projectile motion, perfect for Grades 9-11.

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