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Amir stands on a balcony and throws a ball to his dog, who is at ground level. The ball's height (in meters above the ground), x seconds after Amir threw it, is modeled by: h(x)=-(x-2)^(2)+16 How many seconds after being thrown will the ball hit the ground? seconds

Amir stands on a balcony and throws a ball to his dog, who is at ground level. The ball's height (in meters above the ground), x x seconds after Amir threw it, is modeled by: h(x)=−(x−2)2+16 h(x) = -(x-2)^{2} + 16 How many seconds after being thrown will the ball hit the ground? seconds \text{seconds}

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Q. Amir stands on a balcony and throws a ball to his dog, who is at ground level. The ball's height (in meters above the ground), x x seconds after Amir threw it, is modeled by: h(x)=−(x−2)2+16 h(x) = -(x-2)^{2} + 16 How many seconds after being thrown will the ball hit the ground? seconds \text{seconds}
  1. Understand the problem: We need to find the time when the ball's height is 00, which means when it hits the ground.
  2. Set height equation: Set the height equation equal to zero to find when the ball hits the ground.\newline0=−(x−2)2+160 = -(x-2)^2 + 16
  3. Solve equation for x: First, move the term 1616 to the other side of the equation to isolate the squared term. \newline−(x−2)2=−16-(x-2)^2 = -16 \newline(x−2)2=16(x-2)^2 = 16
  4. Take square root: Take the square root of both sides to solve for xx. \newline(x−2)2=16\sqrt{(x-2)^2} = \sqrt{16} \newlinex−2=±4x - 2 = \pm 4
  5. Solve for x: Solve for x by adding 22 to both possible values of the square root.\newlinex=2+4x = 2 + 4 or x=2−4x = 2 - 4\newlinex=6x = 6 or x=−2x = -2
  6. Interpret results: Since time cannot be negative, we discard the negative value.\newlineThe ball will hit the ground 66 seconds after being thrown.

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