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

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Solve quadratic equations: word problems

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Q. An object is launched from a platform. Its height (in meters),

x

seconds after the launch, is modeled by:

\newline

h(x)=-5x^{2}+20x+60

\newline

How many seconds after launch will the object land on the ground?

\newline

\text{seconds}

Equation setup: We have the equation: $h(x) = -5x^2 + 20x + 60$ $\newline$ To find when the object will land on the ground, we need to find the value of $x$ when $h(x) = 0$ . $\newline$ Set the equation equal to zero: $-5x^2 + 20x + 60 = 0$
Quadratic formula: To solve the quadratic equation $-5x^2 + 20x + 60 = 0$ , we can use the quadratic formula $x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}}$ , where $a = -5$ , $b = 20$ , and $c = 60$ .
Calculate discriminant: First, calculate the discriminant $b^2 - 4ac$ : $\newline$ Discriminant = $20^2 - 4(-5)(60)$ $\newline$ Discriminant = $400 + 1200$ $\newline$ Discriminant = $1600$
Calculate solutions: Since the discriminant is positive, we have two real solutions. Now, calculate the two possible values for $x$ using the quadratic formula: $\newline$ $x = \frac{{-20 \pm \sqrt{1600}}}{{2 \cdot -5}}$ $\newline$ $x = \frac{{-20 \pm 40}}{{-10}}$
Simplify solutions: Calculate the two solutions: $\newline$ First solution: $x = \frac{{-20 + 40}}{{-10}}$ $\newline$ Second solution: $x = \frac{{-20 - 40}}{{-10}}$
Choose physically meaningful solution: Simplify both solutions: $\newline$ First solution: $x = \frac{20}{-10}$ $\newline$ First solution: $x = -2$ (This solution is not physically meaningful, as time cannot be negative.) $\newline$ Second solution: $x = \frac{-60}{-10}$ $\newline$ Second solution: $x = 6$
Choose physically meaningful solution: Simplify both solutions: $\newline$ First solution: $x = \frac{20}{-10}$ $\newline$ First solution: $x = -2$ (This solution is not physically meaningful, as time cannot be negative.) $\newline$ Second solution: $x = \frac{-60}{-10}$ $\newline$ Second solution: $x = 6$ Choose the physically meaningful solution, which is $x = 6$ seconds. This is the time after launch when the object will land on the ground.