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? - Treasure Valley Movers

Maximum Height of a Ball Thrown Upward: A Physics Perspective Using the Quadratic Formula

When a ball is thrown upward with an initial velocity of 49 meters per second, its height over time follows a quadratic trajectory governed by the equation:

\[
h(t) = -4.9t^2 + 49t
\]

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:

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

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

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.