Westonci.ca is the ultimate Q&A platform, offering detailed and reliable answers from a knowledgeable community. Get detailed answers to your questions from a community of experts dedicated to providing accurate information. Experience the ease of finding precise answers to your questions from a knowledgeable community of experts.
Sagot :
Check the picture below, so the ball's path is pretty much like so, and it reaches its hightest at its vertex, so
[tex]~~~~~~\textit{initial velocity in feet} \\\\ h(t) = -16t^2+v_ot+h_o \quad \begin{cases} v_o=\textit{initial velocity}&80\\ \qquad \textit{of the object}\\ h_o=\textit{initial height}&0\\ \qquad \textit{of the object}\\ h=\textit{object's height}&h\\ \qquad \textit{at "t" seconds} \end{cases} \\\\\\ f(t)=80t-16t^2\implies f(t)=-16t^2+80t+0 \\\\[-0.35em] ~\dotfill[/tex]
[tex]\textit{vertex of a vertical parabola, using coefficients} \\\\ y=\stackrel{\stackrel{a}{\downarrow }}{-16}x^2\stackrel{\stackrel{b}{\downarrow }}{+80}x\stackrel{\stackrel{c}{\downarrow }}{+0} \qquad \qquad \left(-\cfrac{ b}{2 a}~~~~ ,~~~~ c-\cfrac{ b^2}{4 a}\right)[/tex]
[tex]\left(-\cfrac{ 80}{2(-16)}~~~~ ,~~~~ 0-\cfrac{ (80)^2}{4(-16)}\right) \implies \left( - \cfrac{ 80 }{ -32 }~~,~~0 - \cfrac{ 6400 }{ -64 } \right) \\\\\\ \left( \cfrac{5}{2}~~,~~100 \right)\implies \underset{~\hfill feet ~~ }{\stackrel{seconds\qquad }{\left( 2\frac{1}{2}~~,~~100 \right)}}[/tex]

We appreciate your visit. Hopefully, the answers you found were beneficial. Don't hesitate to come back for more information. We hope you found what you were looking for. Feel free to revisit us for more answers and updated information. Westonci.ca is your go-to source for reliable answers. Return soon for more expert insights.