Question 1021026
WITHOUT ALGEBRA 2:
{{{h(t)=-16t^2+128t+144}}}
{{{h(t)=-16(t^2+128t/(-16)+144/(-16))}}}
{{{h(t)=-16(t^2-8t-9)}}}
{{{h(t)=-16(t^2-8t+16-16-9)}}}
{{{h(t)=-16((t^2-8t+16)-25)}}}
{{{h(t)=-16((t-4)^2-25)}}}
{{{h(t)=-16(t-4)^2-16*(-25)}}}
{{{h(t)=-16(t-4)^2+400}}}
 
a) From the last equation above we realize that
{{{h(t)<=400}}} ,
so the maximum height of the ball is {{{highlight(400)}}} feet. 
 
b) The ball hits the ground when {{{h(t)=-16t^2+128t+144=0}}} for a {{{t>0}}} ,
so we need a positive solution for {{{-16((t-4)^2-25)=0}}} .
{{{-16((t-4)^2-25)=0}}}<-->{{{(t-4)^2-25=0}}}<-->{{{(t-4)^2=25}}}-->{{{system(t-4=5,"or",t-4=-5)}}} .
{{{t-4=5}}}<-->{{{t=4+5}}}-->{{{t=9}}} is the only option with {{{t>0}}} ,
so the ball hits the ground {{{highlight(9)}}} seconds after it is thrown upwards.

WITH FORMULAS:
a) A quadratic function {{{f(x)=ax^2+bx+c}}} with {{{x<0}}} 
has a maximum for {{{x=(-b)/(2*a)}}} .
 
{{{h(t)=-16t^2+128t+144}}} is such a function with {{{x=t}}} , {{{a=-16<0}}}, and {{{b=128}}} ,
so there is a maximum for
{{{t=(-128)/(2*(-16))=(-128)/(-32)=4}}} , and that maximum is
{{{h(4)=-16*4^2+128*4+144=-16*16+256+144=-256+512+144=highlight(400)}}} .
 
b) The ball hits the ground when {{{h(t)=-16t^2+128t+144=0}}} for a {{{t>0}}} .
So, we are looking for a positive solution to
{{{-16t^2+128t+144=0}}}<-->{{{16t^2-128t-144=0}}}<-->{{{16t^2/16-128t/16-144/16=0}}}<-->{{{t^2-8t-9=0}}} .
 
Since the solutions are {{{t=9}}} and {{{t=-1}}} ,
the ball hits the ground {{{highlight(9)}}} seconds after it is thrown upwards.
 
The equation {{{t^2-8t-9=0}}} can be solved by factoring, by "completing the square", and by using the quadratic formula.
By factoring:
{{{t^2-8t-9=0}}}
{{{(t+1)(t-9)=0}}}--->{{{system(t+1=0,"or",t-9=0)}}}--->{{{system(t=-1,"or",t=9)}}}
By using the quadratic formula:
The quadratic formula says that any solution(s) for
{{{ax^2+bx+c=0}}} , if there is any, are given by
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} ,
so any solution(s) for {{{t^2-8t-9=0}}} ,
with {{{system(t=x,a=1,b=-8,c=-1)}}} , would be given by
{{{t=(-(-8) +- sqrt((-8)^2-4*1*(-9)))/(2*1)=(8 +- sqrt(64+36))/2=(8 +- sqrt(100))/2=(8 +- 10)/2}}}-->{{{system(t=(8+10)/2=18/2=9,"or",t=(8-10)/2=(-2)/2=-1)}}} .
Completing the square is pretty much what was done at the top:
{{{t^2-8t-9=0}}}{{{t^2-8t-9=0}}}<-->{{{t^2-8t+16=9+16}}}<-->{{{(t-4)^2=25}}}-->{{{system(t-4=5,"or",t-4=-5)}}}-->{{{system(t=5+4=9,"or",t=-5+4=-1)}}}