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Ming throws a stone off a bridge into a river below. The stone's height (in meters above the water), x seconds after Ming threw it, is modeled by: h(x)=-5(x-1)^(2)+45 How many seconds after being thrown will the stone reach its maximum height? seconds

Ming throws a stone off a bridge into a river below.\newlineThe stone's height (in meters above the water), xx seconds after Ming threw it, is modeled by:\newlineh(x)=−5(x−1)2+45h(x)=-5(x-1)^{2}+45\newlineHow many seconds after being thrown will the stone reach its maximum height?\newlineseconds\text{seconds}

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Q. Ming throws a stone off a bridge into a river below.\newlineThe stone's height (in meters above the water), xx seconds after Ming threw it, is modeled by:\newlineh(x)=−5(x−1)2+45h(x)=-5(x-1)^{2}+45\newlineHow many seconds after being thrown will the stone reach its maximum height?\newlineseconds\text{seconds}
  1. Given Equation: We are given the equation for the height of the stone above the water as a function of time: h(x)=−5(x−1)2+45 h(x) = -5(x - 1)^2 + 45 . To find the time at which the stone reaches its maximum height, we need to identify the vertex of the parabola represented by this equation. Since the coefficient of the squared term is negative (−5 -5 ), the parabola opens downwards, and the vertex will give us the maximum height.
  2. Identifying the Vertex: The vertex form of a parabola is given by h(x)=a(x−h)2+kh(x) = a(x - h)^2 + k, where (h,k)(h, k) is the vertex of the parabola. In our equation, h(x)=−5(x−1)2+45h(x) = -5(x - 1)^2 + 45, we can see that the vertex is at (h,k)=(1,45)(h, k) = (1, 45). This means that the maximum height is reached at x=1x = 1 second.
  3. Finding the Maximum Height: Since the question asks for the time after being thrown that the stone reaches its maximum height, and we have found that this occurs at x=1x = 1 second, we can conclude that the stone reaches its maximum height 11 second after being thrown.

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