Question 215162
A ball is thrown upward from ground level with an initial velocity of 108 ft per sec. Its height H in feet after T seconds is given by the equation {{{   H= -16T^2 + 108T}}}. At what time will the ball hit the ground?
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When the ball hits the ground, its height above the ground equals zero. 
So just substitute H=0 and  

{{{   H= -16T^2 + 108T}}}

Substituting,

{{{   0= -16T^2 + 108T}}}

Adding {{{16T^2-108T}}} to both sides:

{{{   16T^2 - 108T=0}}}

Factor {{{4T}}} out of the left side:

{{{4T(4T^2-27)=0}}}

Set each factor = 0:

{{{4T=0}}}
{{{T=0/4}}}
{{{T=0}}}

{{{4T^2-27=0}}}
{{{4T^2=27}}}
{{{T^2=27/4}}}
{{{T=sqrt(27/4)}}}
{{{T=sqrt(27)/sqrt(4)}}}
{{{T=sqrt(27)/2}}}
{{{T=sqrt(9*3)/2}}}
{{{T=(sqrt(9)*sqrt(3))/2}}}
{{{T=(3sqrt(3))/2= 2.598076211}}}

which means the ball is on the ground when time is 0,
that is, before the ball first starts upward, and again at when
when it comes back down to the ground 2.6 seconds later.

Edwin</pre>