- Set the height function \(f(x) = -5x^2 + 200\) equal to \(50\) meters.
- Isolate the quadratic term to find \(x^2 = 30\).
- Solve for \(x\) by taking the square root, yielding \(x = \sqrt{30} \approx 5.48\).
- The final answer is \(\boxed{5.48}\).
Explanation
- Problem Analysis
The height of the ball \(x\) seconds after being dropped is given by the quadratic function \(f(x) = -5x^2 + 200\), where \(f(x)\) is the height in meters. We want to determine the time \(x\) at which the ball is exactly \(50\) meters above the ground. Since the ball is dropped, we consider only non-negative values of \(x\). We set the function equal to the target height: \[-5x^2 + 200 = 50.\]
- Isolating the Variable
To solve the equation \(-5x^2 + 200 = 50\) for \(x\), we first isolate the term involving \(x^2\). We subtract \(200\) from both sides of the equation: \[-5x^2 = 50 - 200,\] \[-5x^2 = -150.\] Now, divide both sides by \(-5\): \[x^2 = \frac{-150}{-5},\] \[x^2 = 30.\]
- Finding the Solution
To find \(x\), we take the square root of both sides: \[x = \sqrt{30}.\] Since \(x\) represents time, we take the positive root. Calculating the value of \(\sqrt{30}\), we get approximately \(5.477\). Looking at the provided options, \(5.48\) is the closest value. Thus, the ball is \(50\) meters from the ground after about \(5.48\) seconds.
Examples
This quadratic model demonstrates how physics principles are applied in real-world scenarios. For example, if an engineer is designing a safety net to catch a falling object or testing the trajectory of a sports ball, they use similar equations to predict when an object will reach a specific height. By solving these equations, we can accurately time mechanical responses or evaluate safety requirements in various engineering fields.