A ball is thrown upward with an initial velocity of 49 m/s. Using the formula \(h(t) = -4.9t^2 + 49t\), what is the maximum height reached? - Portal da Acústica

Understanding the Context

This formula models the vertical position \(h(t)\) in meters as a function of time \(t\) in seconds. Because the coefficient of \(t^2\) is negative, the parabola opens downward, meaning the ball eventually rises to a peak and then falls back down — a classic example of projectile motion under gravity (with air resistance neglected).

Finding the Maximum Height

The maximum height occurs at the vertex of the parabola. For any quadratic function in the form \(h(t) = at^2 + bt + c\), the time \(t\) at which the peak height is reached is given by:

\[
t_{\ ext{max}} = -\frac{b}{2a}
\]

Key Insights

In our specific formula, \(a = -4.9\) and \(b = 49\). Substituting these values:

\[
t_{\ ext{max}} = -\frac{49}{2 \ imes (-4.9)} = \frac{49}{9.8} = 5 \ ext{ seconds}
\]

Now substitute \(t = 5\) into the height equation to find the maximum height:

\[
h(5) = -4.9(5)^2 + 49(5) = -4.9(25) + 245 = -122.5 + 245 = 122.5 \ ext{ meters}
\]

Conclusion

Final Thoughts

The maximum height reached by the ball is 122.5 meters. This result illustrates how algebraic expressions rooted in motion physics can precisely predict real-world phenomena, helping students and scientists alike understand the beauty of parabolic motion and optimal performance in projectile dynamics.

Keywords: ball thrown upward, maximum height formula, quadratic motion physics, projectile motion, h(t) = -4.9t² + 49t, vertex of a parabola, maximizing height, physics education.