Question 1154647


What is the maximum height of the ball?

 {{{h(t)=-16t^2+80t+96}}} -> it is a parabola that opens downward, max is at vertex; so, write equation in vertex form


{{{ h(t)=(-16t^2+80t)+96}}}....complete square

 {{{h(t)=-16(t^2+5t)+96}}}

 {{{h(t)=-16(t^2+5t+b^2)-(-16b^2)+96}}}-> {{{b=5/2}}}

 {{{h(t)=-16(t^2+5t+(5/2)^2)+16(5/2)^2+96}}}

 {{{h(t)=-16(t+5/2)^2+16(25/4)+96}}}

 {{{h(t)=-16(t+5/2)^2+4(25)+96}}}

{{{ h(t)=-16(t+5/2)^2+100+96}}}

{{{ h(t)=-16(t+5/2)^2+196}}}

{{{h=5/2}}}, {{{k=196 }}}=> vertex is at ({{{5/2}}},{{{196}}})


How many seconds does it take until the ball hits the ground?


{{{-16t^2+80t+96=0}}}


{{{t=(-80+-sqrt(80^2-4(-16)*96))/(2(-16))}}}


{{{t=(-80+-sqrt(6400+6144))/-32}}}


{{{t=(-80+-sqrt(12544))/-32}}}


{{{t=(-80+-112)/-32}}}


 
{{{t=(-80+112)/-32}}}=>{{{t=-1}}}-> disregard negative solution

{{{t=(-80-112)/-32}}}=>{{{t=6}}}

=> it takes {{{6}}} seconds until the ball hits the ground