Question

A ball is thrown in the air. Its path is modeled by the equation ℎ(t) =− 16t^2 + 56t, where h represents the height of the ball in cm and t represents time in seconds since the ball was thrown. When is the ball at its highest point? What is the highest point?

Answer 1

Check the picture below.

`\textit{vertex of a vertical parabola, using coefficients} \\\\ h(t)=\stackrel{\stackrel{a}{\downarrow }}{-16}t^2\stackrel{\stackrel{b}{\downarrow }}{+56}t\stackrel{\stackrel{c}{\downarrow }}{+0} \qquad \qquad \left(-\cfrac{ b}{2 a}~~~~ ,~~~~ c-\cfrac{ b^2}{4 a}\right)`

`\left(-\cfrac{ 56}{2(-16)}~~~~ ,~~~~ 0-\cfrac{ (56)^2}{4(-16)}\right) \implies \left( - \cfrac{ 56 }{ -32 }~~,~~0 - \cfrac{ 3136 }{ -64 } \right) \\\\\\ \left( \cfrac{7}{4}~~,~~49 \right)\implies \stackrel{seconds\qquad feet}{\left(1(3)/(4) ~~ ~~,~ ~~ ~49 \right)}`