Question 348238
You can use the equation to get clues for how to proceed
Like lots of motion problems, it starts at {{{t = 0}}}
What is {{{h}}} at {{{t = 0}}}?
{{{ h=16t^2 + 48t + 160}}}
{{{ h=16*0^2 + 48*0 + 160}}}
{{{h = 160}}}
This is saying that you are 160 ft above ground when you 
throw the object up. There will be some value of {{{t}}}
that will give me {{{h =160}}} when the object is on it's way
down. This means it's back at your level on top of the building.
You want to find {{{t}}} at {{{h = 160 - 160}}} which is {{{h = 0}}}
{{{ h= 16t^2 + 48t + 160}}}
{{{ 0= 16t^2 + 48t + 160}}}
{{{ 0= t^2 + 3t + 10}}}
I can solve using quadratic formula
{{{t = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
{{{a = 1}}}
{{{b = 3}}}
{{{c = 10}}}
{{{t = (-3 +- sqrt( 3^2-4*1*10 ))/(2*1) }}}
{{{t = (-3 +- sqrt( 9 - 40 ))/2 }}}
This makes no sense- the square root is of a negative
The equation should be
{{{ h = -16^2 + 48t + 160}}}, then
{{{ 0 = -t^2 + 3t + 10}}}
{{{a = -1}}}
{{{b = 3}}}
{{{c = 10}}}
{{{t = (-3 +- sqrt( 3^2-4*(-1)*10 ))/(2*(-1)) }}}
{{{t = (-3 +- sqrt( 9 + 40 ))/-2 }}}
{{{t = (-3 + 7)/-2}}} (can't use this - negative number)
{{{t = (-3 -7)/-2}}}
{{{t = 5}}}
The object hits the ground in 5 sec
I can plot this as {{{h}}} vs {{{t}}} (note that {{{h}}} is divided by {{{16}}} here) 
{{{ graph( 400, 400, -1, 10, -1, 13, -x^2 + 3x + 10) }}}