Question

A ball is kicked upward with an initial velocity of 52 feet per second. The ball's height, h (in feet), from the ground Is

modeled by h= -16t^2 +52t where t is measured in seconds. How much time does the ball take to reach its highest
point? What is its height at this 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 }}{+52}t\stackrel{\stackrel{c}{\downarrow }}{+0} \qquad \qquad \left(-\cfrac{ b}{2 a}~~~~ ,~~~~ c-\cfrac{ b^2}{4 a}\right) \\\\\\ \left(-\cfrac{ 52}{2(-16)}~~~~ ,~~~~ 0-\cfrac{ (52)^2}{4(-16)}\right) \implies \left( - \cfrac{ 52 }{ -32 }~~,~~0 - \cfrac{ 2704 }{ -64 } \right)`

`\left( \cfrac{13}{8}~~,~~\cfrac{169}{4} \right)\implies \underset{\stackrel{\uparrow \hspace{3em}}{seconds\qquad }}{\stackrel{\stackrel{\qquad feet}{\hspace{3em}\downarrow }}{\left( 1(5)/(8)~~,~~42(1)/(4) \right)}}`