QuestionJuly 21, 2025

A ball is thrown vertically upward. After r seconds, it's height him feet) is given by the function h(t)=40t-16t^2 After how long will It reach its maximum height? Do not round your answer. Time: square seconds

A ball is thrown vertically upward. After r seconds, it's height him feet) is given by the function h(t)=40t-16t^2 After how long will It reach its maximum height? Do not round your answer. Time: square seconds
A ball is thrown vertically upward. After r seconds, it's height him feet) is given by the function h(t)=40t-16t^2 After how long will It reach its maximum height? Do not round your answer. Time: square seconds

Solution
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Answer

\frac{5}{4} seconds Explanation 1. Identify the vertex of the parabola The function h(t) = 40t - 16t^2 is a quadratic equation in the form at^2 + bt + c. The maximum height is at the vertex. For ax^2 + bx + c, the time to reach the maximum height is given by **t = -\frac{b}{2a}**. 2. Calculate the time for maximum height Here, a = -16 and b = 40. Substitute these values into the formula: t = -\frac{40}{2 \times (-16)} = \frac{40}{32} = \frac{5}{4}.

Explanation

1. Identify the vertex of the parabola<br /> The function $h(t) = 40t - 16t^2$ is a quadratic equation in the form $at^2 + bt + c$. The maximum height is at the vertex. For $ax^2 + bx + c$, the time to reach the maximum height is given by **$t = -\frac{b}{2a}$**.<br /><br />2. Calculate the time for maximum height<br /> Here, $a = -16$ and $b = 40$. Substitute these values into the formula: $t = -\frac{40}{2 \times (-16)} = \frac{40}{32} = \frac{5}{4}$.
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