An object is launched from a platform. Its height (in meters), x seconds after the launch, is modeled by h(x)=-5(x+1)(x-9) How many seconds after launch will the object hit the ground? ◻ seconds

Set Equation Equal: To find when the object hits the ground, we need to solve for x when h(x) = 0.Simplify Equation: Set the equation h(x) = -5(x+1)(x-9) equal to 0 and solve for x. 0 = -5(x+1)(x-9)Quadratic Equation: Divide both sides by -5 to simplify the equation. 0 = (x+1)(x-9)Solve for x: Now we have a quadratic equation that can be solved by setting each factor equal to zero. x+1 = 0 or x-9 = 0Discard Negative Solution: Solve each equation for x. x = -1 or x = 9Final Time Calculation: Since time can't be negative, we discard x = -1 and take x = 9 as the time when the object hits the ground.

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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)=-5(x+1)(x-9)
How many seconds after launch will the object hit the ground?

◻ seconds

An object is launched from a platform. $\newline$ Its height (in meters), $x$ seconds after the launch, is modeled by $\newline$ $h(x)=-5(x+1)(x-9)$ $\newline$ How many seconds after launch will the object hit the ground? $\newline$ $\square$ seconds

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Q. An object is launched from a platform.

\newline

Its height (in meters),

x

seconds after the launch, is modeled by

\newline

h(x)=-5(x+1)(x-9)

\newline

How many seconds after launch will the object hit the ground?

\newline

\square

seconds

Set Equation Equal: To find when the object hits the ground, we need to solve for $x$ when $h(x) = 0$ .
Simplify Equation: Set the equation $h(x) = -5(x+1)(x-9)$ equal to $0$ and solve for $x$ . $0 = -5(x+1)(x-9)$
Quadratic Equation: Divide both sides by $-5$ to simplify the equation. $\newline$ $0 = (x+1)(x-9)$
Solve for x: Now we have a quadratic equation that can be solved by setting each factor equal to zero. $\newline$ $x+1 = 0$ or $x-9 = 0$
Discard Negative Solution: Solve each equation for $x$ . $x = -1$ or $x = 9$
Final Time Calculation: Since time can't be negative, we discard $x = -1$ and take $x = 9$ as the time when the object hits the ground.