A ball is thrown upward with an equation h(t)=-4t²+8t, where h is the height in meters and t is the time in seconds. What is the maximum height that a ball can reach?
Answer (1)
[tex]\begin{gathered}\begin{gathered}{\underline{\huge \mathbb{A} {\large \mathrm {NSWER : }}}} \\\end{gathered}\end{gathered}[/tex]To find the maximum height that the ball can reach using the given equation [tex]h(t) = -4t^2 + 8t[/tex], we will follow these steps: Identify the Quadratic Function The equation [tex]h(t) = -4t^2 + 8t[/tex] is a quadratic function in the standard form [tex]h(t) = at^2 + bt + c[/tex], where: [tex]a = -4[/tex] [tex]b = 8[/tex] [tex]c = 0[/tex] (since there is no constant term) Find the Vertex of the Quadratic Function The maximum height of a parabola that opens downward (which is the case here since [tex]a < 0[/tex]) occurs at the vertex. The time at which the maximum height occurs can be found using the formula: [tex]t = -\frac{b}{2a}[/tex] Substituting the values of [tex]a[/tex] and [tex]b[/tex]: [tex]t = -\frac{8}{2 \cdot -4} = \frac{8}{8} = 1 \text{ s}[/tex] Calculate the Maximum Height Now, substitute [tex]t = 1[/tex] back into the original height equation to find the maximum height: [tex]h(1) = -4(1)^2 + 8(1)[/tex] Calculating this: [tex]h(1) = -4 + 8 = 4 \text{ meters}[/tex] Conclusion The maximum height that the ball can reach is 4 meters.[tex]\sf\color{green}{⊱⋅ ────────────────────── ⋅⊰}[/tex][tex]\begin{gathered} \boxed{\begin{array}{cc} \sf \footnotesize \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: ···\: ʚ\: \: \: \: \: \: Hope\:it\:helps \: \: \: \: \: ɞ \:··· \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \sf\footnotesize\:\# CarryOnLearning \\ \sf\footnotesize ૮₍´˶ \: • \: . \: • \: ⑅ ₎ა \: \leadsto \footnotesize\sf\color{purple} \underline{Study\:Well!}\end{array}}\end{gathered}[/tex][tex]\sf\color{green}{⊱⋅ ────────────────────── ⋅⊰}[/tex][tex]\large\qquad\qquad\qquad\tt MARCH/14/2025 [/tex]
