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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), xx seconds after Amir threw it, is modeled by: h(x)=−(x−2)2+16h(x)=-(x-2)^2 +16. What is the height of the ball at the time it is thrown.

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Full solution

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), xx seconds after Amir threw it, is modeled by: h(x)=−(x−2)2+16h(x)=-(x-2)^2 +16. What is the height of the ball at the time it is thrown.
  1. Identify initial time: Identify the time at which the ball is thrown.\newlineThe time at which the ball is thrown is the initial time, which is x=0x = 0 seconds.
  2. Substitute into equation: Substitute the initial time into the height equation.\newlineWe need to substitute x=0x = 0 into the height equation h(x)=−(x−2)2+16h(x) = -(x-2)^2 + 16 to find the height of the ball at the time it is thrown.
  3. Calculate height: Perform the substitution and calculate the height.\newlineh(0)=−(0−2)2+16h(0) = - (0 - 2)^2 + 16\newlineh(0)=−(2)2+16h(0) = - (2)^2 + 16\newlineh(0)=−4+16h(0) = - 4 + 16\newlineh(0)=12h(0) = 12 meters

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