question a rocket is launched from a tower. the height of the rocket, y in feet, is related to the time…

question a rocket is launched from a tower. the height of the rocket, y in feet, is related to the time after launch, x in seconds, by the given equation. using this equation, find the maximum height reached by the rocket, to the nearest tenth of a foot. y = -16x² + 124x + 110

question a rocket is launched from a tower. the height of the rocket, y in feet, is related to the time after launch, x in seconds, by the given equation. using this equation, find the maximum height reached by the rocket, to the nearest tenth of a foot. y = -16x² + 124x + 110

Answer

Explanation:

Step1: Identify the coefficients

The quadratic - function is in the form $y = ax^{2}+bx + c$, where $a=-16$, $b = 124$, and $c = 110$.

Step2: Find the x - value of the vertex

The x - value of the vertex of a quadratic function $y = ax^{2}+bx + c$ is given by $x=-\frac{b}{2a}$. Substitute $a=-16$ and $b = 124$ into the formula: $x=-\frac{124}{2\times(-16)}=\frac{124}{32}=\frac{31}{8}=3.875$.

Step3: Find the y - value of the vertex

Substitute $x = 3.875$ into the equation $y=-16x^{2}+124x + 110$. $y=-16\times(3.875)^{2}+124\times3.875 + 110$. First, calculate $(3.875)^{2}=15.015625$. Then, $-16\times(3.875)^{2}=-16\times15.015625=-240.25$. $124\times3.875 = 478.5$. $y=-240.25+478.5 + 110$. $y=348.25\approx348.3$.

Answer:

$348.3$