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An object is launched from a platform. Its height (in meters), x seconds after the launch, is modeled by: h(x)=-5x^(2)+20 x+60 How many seconds after launch will the object land on the ground? seconds

An object is launched from a platform. Its height (in meters), xx seconds after the launch, is modeled by:\newlineh(x)=−5x2+20x+60h(x)=-5x^{2}+20x+60\newlineHow many seconds after launch will the object land on the ground?\newlineseconds\text{seconds}

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Q. An object is launched from a platform. Its height (in meters), xx seconds after the launch, is modeled by:\newlineh(x)=−5x2+20x+60h(x)=-5x^{2}+20x+60\newlineHow many seconds after launch will the object land on the ground?\newlineseconds\text{seconds}
  1. Equation setup: We have the equation: h(x)=−5x2+20x+60 h(x) = -5x^2 + 20x + 60 \newlineTo find when the object will land on the ground, we need to find the value of x x when h(x)=0 h(x) = 0 .\newlineSet the equation equal to zero: −5x2+20x+60=0 -5x^2 + 20x + 60 = 0
  2. Quadratic formula: To solve the quadratic equation −5x2+20x+60=0-5x^2 + 20x + 60 = 0, we can use the quadratic formula x=−b±b2−4ac2ax = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}}, where a=−5a = -5, b=20b = 20, and c=60c = 60.
  3. Calculate discriminant: First, calculate the discriminant b2−4acb^2 - 4ac: \newlineDiscriminant = 202−4(−5)(60)20^2 - 4(-5)(60)\newlineDiscriminant = 400+1200400 + 1200\newlineDiscriminant = 16001600
  4. Calculate solutions: Since the discriminant is positive, we have two real solutions. Now, calculate the two possible values for xx using the quadratic formula:\newlinex=−20±16002⋅−5x = \frac{{-20 \pm \sqrt{1600}}}{{2 \cdot -5}}\newlinex=−20±40−10x = \frac{{-20 \pm 40}}{{-10}}
  5. Simplify solutions: Calculate the two solutions:\newlineFirst solution: x=−20+40−10x = \frac{{-20 + 40}}{{-10}}\newlineSecond solution: x=−20−40−10x = \frac{{-20 - 40}}{{-10}}
  6. Choose physically meaningful solution: Simplify both solutions:\newlineFirst solution: x=20−10x = \frac{20}{-10}\newlineFirst solution: x=−2x = -2 (This solution is not physically meaningful, as time cannot be negative.)\newlineSecond solution: x=−60−10x = \frac{-60}{-10}\newlineSecond solution: x=6x = 6
  7. Choose physically meaningful solution: Simplify both solutions:\newlineFirst solution: x=20−10 x = \frac{20}{-10} \newlineFirst solution: x=−2 x = -2 (This solution is not physically meaningful, as time cannot be negative.)\newlineSecond solution: x=−60−10 x = \frac{-60}{-10} \newlineSecond solution: x=6 x = 6 Choose the physically meaningful solution, which is x=6 x = 6 seconds. This is the time after launch when the object will land on the ground.

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