Set \(h(t) = 0\): \(-5t^2 + 20t + 10 = 0\).
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- Substitute into second equation: \(4(3) - y = 10\) â \(12 - y = 10\) â \(y = 2\).
- So \(x + y = 3 + 2 = 5\).
- A ball is thrown upward with initial velocity 20 m/s from a height of 10 m. Its height at time \(t\) is \(h(t) = -5t^2 + 20t + 10\). When does it hit the ground?
- Divide by -5: \(t^2 - 4t - 2 = 0\).
- Use quadratic formula: \(t = rac{4 \pm \sqrt{(-4)^2 - 4(1)(-2)}}{2} = rac{4 \pm \sqrt{16 + 8}}{2} = rac{4 \pm \sqrt{24}}{2} = rac{4 \pm 2\sqrt{6}}{2} = 2 \pm \sqrt{6}\).
- Take positive root: \(t = 2 + \sqrt{6} pprox 4.45\) seconds.
