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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. \newlineIts height (in meters), xx seconds after the launch, is modeled by\newlineh(x)=−5(x+1)(x−9)h(x)=-5(x+1)(x-9)\newlineHow many seconds after launch will the object hit the ground?\newline□\square seconds

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Full solution

Q. An object is launched from a platform. \newlineIts height (in meters), xx seconds after the launch, is modeled by\newlineh(x)=−5(x+1)(x−9)h(x)=-5(x+1)(x-9)\newlineHow many seconds after launch will the object hit the ground?\newline□\square seconds
  1. Set Equation Equal: To find when the object hits the ground, we need to solve for xx when h(x)=0h(x) = 0.
  2. Simplify Equation: Set the equation h(x)=−5(x+1)(x−9)h(x) = -5(x+1)(x-9) equal to 00 and solve for xx.0=−5(x+1)(x−9)0 = -5(x+1)(x-9)
  3. Quadratic Equation: Divide both sides by −5-5 to simplify the equation.\newline0=(x+1)(x−9)0 = (x+1)(x-9)
  4. Solve for x: Now we have a quadratic equation that can be solved by setting each factor equal to zero.\newlinex+1=0x+1 = 0 or x−9=0x-9 = 0
  5. Discard Negative Solution: Solve each equation for xx.x=−1x = -1 or x=9x = 9
  6. Final Time Calculation: Since time can't be negative, we discard x=−1x = -1 and take x=9x = 9 as the time when the object hits the ground.

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