Question 1158667

{{{H=-16t^2+38t+101}}}

1) maximum

you have a parabola, upside down, and max is at vertex

so, write your equation in vertex form

{{{H=(-16t^2+38t)+101}}}.......factor out {{{-16}}}

{{{H=-16(t^2-38t/16)+101}}}

{{{H=-16(t^2-19t/8)+101}}}

{{{H=-16(t^2-(19/8)+b^2)-(-16)b^2+101}}}.........{{{b=(19/8)/2=19/16}}}

{{{H=-16(t^2-19t/8+(19/16)^2)+16(19/16)^2+101}}}

{{{H=-16(t-19/16)^2+361/16+101}}}

{{{H=-16(t-19/16)^2+1977/16}}}


=> vertex is at({{{ 19/16}}}, {{{1977/16}}})=( {{{1.19}}}, {{{123.56}}})


so, the ball will reach max in {{{t=1.19}}} seconds


2). 
the max height of the ball is {{{123.56ft}}}


3.

the ball will return to the ground when {{{H=0}}}

{{{0=-16t^2+38t+101}}}..........solve for {{{t}}} using quadratic formula


{{{t=(-b+-sqrt(b^2-4ac))/2a}}}


since {{{a=-16}}},{{{ b=38}}}, and {{{c=101}}} we have


{{{t=(-38+-sqrt(38^2-4(-16)*101))/(2(-16))}}}


{{{t=(-38+-sqrt(1444+6464))/(-32)}}}


{{{t=(-38+-sqrt(7908))/(-32)}}}


{{{t=(-38+-88.93)/(-32)}}}


solutions:

{{{t=(-38+88.93)/(-32)=-1.59}}}

or

{{{t=(-38-88.93)/(-32)=3.97}}}


disregard negative solution

so, the ball will return to the ground in {{{3.97}}} seconds