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

Define Ground Impact: To find when the object will hit the ground, we need to determine when the height h(x) is equal to zero. This is because the height of the object will be zero when it touches the ground.Set Height Function: Set the height function h(x) equal to zero and solve for x. 0 = -5(x+1)(x-9)Solve for x: Since the equation is already factored, we can set each factor equal to zero and solve for x. First, set the first factor equal to zero: -5(x+1) = 0Discard Negative Solution: Divide both sides of the equation by -5 to isolate the term with x. (x+1) = 0Final Result: Subtract 1 from both sides to solve for x. x = -1Now, set the second factor equal to zero: (x-9) = 0Add 9 to both sides to solve for x. x = 9We have two solutions for x: x = -1 and x = 9. However, since time cannot be negative in this context, we discard the negative value. The object will hit the ground 9 seconds after launch.

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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? $\newline$ $\text{seconds}$

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Algebra 2

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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

\text{seconds}

Define Ground Impact: To find when the object will hit the ground, we need to determine when the height $h(x)$ is equal to zero. This is because the height of the object will be zero when it touches the ground.
Set Height Function: Set the height function $h(x)$ equal to zero and solve for $x$ . $0 = -5(x+1)(x-9)$
Solve for x: Since the equation is already factored, we can set each factor equal to zero and solve for x. $\newline$ First, set the first factor equal to zero: $\newline$ $-5(x+1) = 0$
Discard Negative Solution: Divide both sides of the equation by $-5$ to isolate the term with $x$ . $\newline$ $(x+1) = 0$
Final Result: Subtract $1$ from both sides to solve for $x$ . $\newline$ $x = -1$
Final Result: Subtract $1$ from both sides to solve for $x$ . $\newline$ $x = -1$ Now, set the second factor equal to zero: $\newline$ $(x-9) = 0$
Final Result: Subtract $1$ from both sides to solve for $x$ . $\newline$ $x = -1$ Now, set the second factor equal to zero: $\newline$ $(x-9) = 0$ Add $9$ to both sides to solve for $x$ . $\newline$ $x = 9$
Final Result: Subtract $1$ from both sides to solve for $x$ . $\newline$ $x = -1$ Now, set the second factor equal to zero: $\newline$ $(x-9) = 0$ Add $9$ to both sides to solve for $x$ . $\newline$ $x = 9$ We have two solutions for $x$ : $x = -1$ and $x = 9$ . However, since time cannot be negative in this context, we discard the negative value. The object will hit the ground $9$ seconds after launch.