A ball is thrown upward with an initial velocity of 20 m/s. Using \(h = -5t^2 + 20t\), find the maximum height reached.
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- Solve using the quadratic formula: \(x = \frac{-70 \pm \sqrt{4900 + 5600}}{8}\).
- Approximate: \(x \approx \frac{-70 + 103.92}{8} = 4.49\) meters.
- #### 4.49
- The height function is \(h(t) = -5t^2 + 20t\).
- The vertex (maximum) occurs at \(t = -\frac{b}{2a} = -\frac{20}{2 \times -5} = 2\) seconds.
- Substitute \(t = 2\) into the height equation: \(h(2) = -5(2)^2 + 20(2) = -20 + 40 = 20\) meters.
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