Question 1131808

{{{h(t) = -t^2 + 96t + 256}}}


(a) Find the maximum height of the projectile._______ft

{{{h(t) = -16t^2 + 96t + 256}}}-> the maximum is at vertex
{{{h(t) = -16(t^2 + 6t) + 256}}}...complete square
{{{h(t) = -16(t^2 + 6t+b^2)-(-16)b^2 + 256}}}
{{{h(t) = -16(t^2 + 6t+3^2)+16*3^2 + 256}}}
{{{h(t) = -16(t+3)^2+144 + 256}}}
{{{h(t) = -16(t+3)^2+400}}}

=> {{{h=-3}}} and {{{k=400}}}

so, vertex is at ({{{-3}}},{{{400}}})

the maximum height of the projectile._{{{400}}}__ft



(b) Find the time {{{t }}}when the projectile achieves its maximum height.

The{{{ t}}} value is {{{-b/2a=-96/-32}}} or {{{t=3}}}seconds
 


(c) Find the time {{{t }}}when the projectile has a height of {{{0}}} feet.

{{{-16t^2+96t+256=0}}}......divide by {{{-16}}}
{{{t^2-6t-16=0}}}
{{{(t-8)(t+2)=0}}}
{{{t=8}}} seconds, (need only positive root )