Question

If the ball leaves her hand 6 feet above the ground at 65 feet/second: a) what is the maximum height of the ball

Answer 1

$~~~~~~\textit{initial velocity in feet} \\\\ h(t) = -16t^2+v_ot+h_o \quad \begin{cases} v_o=\textit{initial velocity}&65\\ \qquad \textit{of the object}\\ h_o=\textit{initial height}&6\\ \qquad \textit{of the object}\\ h=\textit{object's height}&\\ \qquad \textit{at "t" seconds} \end{cases} \\\\\\ h(t)=-16t^2+65t+6$

Check the picture below, the maximum occurs at the vertex of the parabolic path, so

$\textit{vertex of a vertical parabola, using coefficients} \\\\ h(t)=\stackrel{\stackrel{a}{\downarrow }}{-16}t^2\stackrel{\stackrel{b}{\downarrow }}{+65}t\stackrel{\stackrel{c}{\downarrow }}{+6} \qquad \qquad \left(-\cfrac{ b}{2 a}~~~~ ,~~~~ c-\cfrac{ b^2}{4 a}\right)$

$\left(-\cfrac{ 65}{2(-16)}~~~~ ,~~~~ 6-\cfrac{ (65)^2}{4(-16)}\right)\implies \left(\cfrac{65}{32}~~,~~6+\cfrac{4225}{32} \right) \\\\\\ \left(\cfrac{65}{32}~~,~~\cfrac{4609}{32} \right) \implies \stackrel{maximum~height}{(2.03125~~,~~\stackrel{\downarrow }{72.015625})}$