A ball is done the air with an upward velocity of 32 feet a second it’s height H in fee after T seconds is given by the function h=-16t^2+32T +6. A. In how many seconds does the ball reach its maximum height? Round to the nearest hundredth if necessary.
B. What is the balls maximum height?

A. 1s;22 ft
B. 2s;22 ft
C. 1s;54 ft
D. 2s;6ft

Respuesta :

check the picture below.

so the ball reaches a maximum height at its vertex,

[tex] \bf \textit{vertex of a vertical parabola, using coefficients}
\\\\
h=\stackrel{\stackrel{a}{\downarrow }}{-16}t^2\stackrel{\stackrel{b}{\downarrow }}{+32}t\stackrel{\stackrel{c}{\downarrow }}{+6}
\qquad \qquad
\left(-\cfrac{ b}{2 a}~~~~ ,~~~~ c-\cfrac{ b^2}{4 a}\right)
\\\\\\
\left(-\cfrac{32}{2(-16)}~~,~~6-\cfrac{32^2}{4(-16)} \right)\implies \left(\cfrac{32}{32}~~,~~6-\cfrac{1024}{-64} \right)
\\\\\\
(1~~,~~6+16)\implies (\stackrel{seconds}{1}~,~\stackrel{\stackrel{feet}{high}}{22}) [/tex]

Ver imagen jdoe0001