Question 1170723
a. About how long is the ball in the air?

{{{g(t)=0}}}

{{{0=-16t^2+120t+4}}}

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

{{{t=(-120+-sqrt(120^2-4(-16)4))/(2(-16))}}}....

{{{t=(-120+-sqrt(14656))/(-32)}}}

{{{t=(-120+-121.06)/(-32)}}}

need only positive solution

{{{t=(-120-121.06)/(-32)}}}

{{{t}}} ≅ {{{7.53}}}seconds


b. When is the ball back to {{{4}}} feet off the ground?

{{{g(t)=4}}}

{{{4=-16t^2+120t+4}}}....simplify

{{{1=-4t^2+30t+1}}}

{{{0=-4t^2+30t+1-1}}}

{{{0=-4t^2+30t}}}

{{{0=-4t(t-30/4)}}}

{{{0=-4t(t-15/2)}}}

need only positive solution

{{{0=(t-15/2)}}}

{{{t=15/2}}}

{{{t=7.5}}}seconds


c. How high is the ball after {{{4}}} seconds?

{{{t=4}}}

{{{g(4)=-16*4^2+120*4+4}}}

{{{g(4)=228}}} feet


d. When is the ball {{{150}}} feet in the air?

{{{g(t)=150}}}

{{{150=-16t^2+120t+4}}}

{{{0=-16t^2+120t+4-150}}}

{{{0=-16t^2+120t-146}}}

{{{t=(-120+-sqrt(120^2-4(-16)(-146)))/(2(-16))}}}

{{{t=(-120+-sqrt(5056))/(-32)}}}

need only positive solution

{{{t=(-120+-71.1)/(-32)}}}

{{{t=(-120-71.1)/(-32)}}}

{{{t=5.971875}}}

{{{t}}} ≅ {{{5.97}}}seconds


e. What is the maximum height the ball reaches? How many seconds does it take for this max to occur?

max is at vertex, write equation in vertex form

{{{g(x)=-16t^2+120t+4}}}....complete square

{{{g(x)=(-16t^2+120t)+4}}}

{{{g(x)=-16(t^2-(15/2)t)+4}}}

{{{g(x)=-16(t^2-(15/2)t+b^2)-(-16b^2)+4}}}.......{{{b=(15/2)/2=15/4}}}

{{{g(x)=-16(t^2-(15/2)t+(15/4)^2)-(-16(15/4)^2)+4}}}

{{{g(x)=-16(t-15/4)^2+16(225/16)+4}}}

{{{g(x)=-16(t-15/4)^2+225+4}}}

{{{g(x)=-16(t-15/4)^2+229}}}

=>{{{h=15/4}}} and {{{k=229}}}

vetex is at ({{{15/4}}},{{{229}}})

max is {{{229}}}, and it will take {{{15/4}}} seconds for this max to occur