Question 882158
{{{h(t)= -5(t-1.4)^2 +10}}} is the height of the ball above the ground {{{t}}} seconds after it is thrown.
{{{h(t)=0}}} means the ball is 0 meters above the ground, meaning that it is on the ground.
The equation making {{{h(t)=0}}} gives you 2 answers for {{{t}}} , one negative and one positive.
The negative answer makes no sense, because we only care about the time after the ball was thrown.
(We do not know or care where the ball was for the "negative" values of {{{t}}} that represent before the throw. Maybe it was in someone's pocket before being thrown, but we do not care).
 
{{{system(h(t)= -5(t-1.4)^2 +10,h(t)=0)}}} ---> {{{-5(t-1.4)^2 +10=0}}} ---> {{{10=5(t-1.4)^2}}} ---> {{{10/5=(t-1.4)^2}}} ---> {{{2=(t-1.4)^2}}}
From {{{(t-1.4)^2=2}}} we get the two answers:
{{{(t-1.4)^2=2}}}--->{{{system(t-1.4=sqrt(2),"or",t-1.4=sqrt(2))}}}--->{{{system(t=sqrt(2)+1.4,"or",t=-sqrt(2)+1.4)}}}
{{{highlight(t=sqrt(2)+1.4=about2.8)}}} is the answer you want.
The ball hits the ground {{{highlight(about2.8)}}} seconds after being thrown.