- Define the height function as \(h(t) = -16t^2 + 120t\).
- Set \(h(t) = 200\) and rearrange the equation into the quadratic form \(2t^2 - 15t + 25 = 0\).
- Use the quadratic formula to solve for \(t\), yielding \(t = 2.5 \text{ s}\) and \(t = 5.0 \text{ s}\).
- Select the matching option from the list, which is \(\boxed{2.5 \text{ s}}\).
Explanation
- Problem Analysis
The problem asks for the time \(t\) in seconds after which a projectile reaches a height of \(200 \text{ ft}\), given an initial velocity \(v = 120 \text{ ft/s}\) and an acceleration \(a = -16 \text{ ft/s}^2\). The height function is given by \(h(t) = at^2 + vt + h_0\). Since the object is launched from the ground, the initial height \(h_0\) is \(0\). Thus, the function is \(h(t) = -16t^2 + 120t\). We need to solve for \(t\) such that \(h(t) = 200\).
- Setting up the Quadratic Equation
To find the time \(t\), we set the height function equal to \(200 \text{ ft}\):
\[-16t^2 + 120t = 200\]
Rearranging this into the standard quadratic form \(ax^2 + bx + c = 0\), we get:
\[16t^2 - 120t + 200 = 0\]
To simplify the equation, we can divide all terms by \(8\):
\[2t^2 - 15t + 25 = 0\]
This is a quadratic equation that can be solved using the quadratic formula \(t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). Here, \(a = 2\), \(b = -15\), and \(c = 25\).
- Solving the Equation
Applying the quadratic formula:
\[t = \frac{-(-15) \pm \sqrt{(-15)^2 - 4(2)(25)}}{2(2)}\]
\[t = \frac{15 \pm \sqrt{225 - 200}}{4}\]
\[t = \frac{15 \pm \sqrt{25}}{4}\]
\[t = \frac{15 \pm 5}{4}\]
This gives us two possible times:
\[t_1 = \frac{15 + 5}{4} = \frac{20}{4} = 5.0 \text{ s}\]
\[t_2 = \frac{15 - 5}{4} = \frac{10}{4} = 2.5 \text{ s}\]
The object reaches \(200 \text{ ft}\) on its way up at \(2.5 \text{ seconds}\) and again on its way down at \(5.0 \text{ seconds}\). Comparing this with the given options, \(2.5 \text{ s}\) is one of the answers provided.
- Conclusion
The object reaches the height of \(200 \text{ ft}\) at \(t = 2.5 \text{ s}\) and \(t = 5.0 \text{ s}\). Among the given options, \(b. 2.5 \text{ s}\) is the correct choice. Therefore, the projectile reaches the height of \(200 \text{ ft}\) after \(2.5 \text{ seconds}\) during its ascent. The final answer is \(\boxed{2.5 \text{ s}}\).
Examples
This mathematical model of projectile motion is essential in fields like sports science and ballistics. For example, coaches use these calculations to analyze the arc of a basketball shot or a soccer ball's trajectory to optimize player performance. By understanding when an object reaches a certain height, professionals can determine the ideal launch angle and velocity to ensure the projectile hits its target.