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The problem provides the height function \( h(t) = -5t^2 + 26t + 1.4 \), which represents the height of a ball kicked during a soccer game, where \( h(t) \) is the height in meters and \( t \) is the time in seconds. We are tasked with finding the time at which the ball reaches its maximum height.

This is a quadratic function of the form:

\[
h(t) = at^2 + bt + c
\]

In this case:
- \( a = -5 \)
- \( b = 26 \)
- \( c = 1.4 \)

To find the time at which the maximum height occurs, we use the formula for the vertex of a quadratic function:

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

Substituting the given values for \( a \) and \( b \):

\[
t = \frac{-26}{2(-5)} = \frac{-26}{-10} = 2.6 \, \text{seconds}
\]

Thus, the ball reaches its maximum height at **2.6 seconds**.

Would you like to know the maximum height, or any additional details?

Here are some related questions:
1. What is the maximum height the ball reaches?
2. How can we calculate the time when the ball hits the ground?
3. How would the equation change if the ball were kicked with a different initial velocity?
4. What are the effects of air resistance on the height of the ball?
5. How do we derive the formula for the time of maximum height for a quadratic function?

**Tip:** To find the maximum or minimum of any quadratic function, the vertex formula \( t = \frac{-b}{2a} \) is very useful, especially when analyzing real-world applications like projectile motion.

Durante un partido de fútbol, un jugador le da un puntapié a la pelota, tal que la altura está expresada por la función h(t) = -5t^2 + 26t + 1.4, donde h es la altura en metros y t es el tiempo en segundos. ¿En qué tiempo la pelota alcanza su máxima altura?

Learn how to calculate the time when a soccer ball reaches its maximum height using the quadratic function h(t) = -5t^2 + 26t + 1.4. This step-by-step solution demonstrates how to apply the vertex formula for a quadratic equation, suitable for Grades 9-12.

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