Question 348691
{{{-32t^2 + 150t + 5 = h}}}
{{{h}}} is the height above ground.
At {{{t=0}}} the height is {{{5}}} ft
as it should be.
At {{{t = 1}}} sec
{{{-32*1^2 + 150*1 + 5 = 123}}} ft ( it's going up)
At {{{t = 2}}} sec
{{{-32*2^2 + 150*2 + 5 = 177}}} ft (still going up)
At {{{t = 3}}} sec
{{{-32*3^2 + 150*3 + 5 = 167}}} ft (it's coming down)
It's going to hit ground sometime after {{{3}}} sec
Make {{{h = 0}}} (ground)
{{{-32t^2 + 150t + 5 = 0}}}
Solve using quadratic formula
 {{{t = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
{{{a = -32}}}
{{{b = 150}}}
{{{c = 5}}}
 {{{t = (-150 +- sqrt( 150^2-4*(-32)*5 ))/(2*(-32)) }}}
 {{{t = (-150 +- sqrt( 23140 ))/(-64) }}}
{{{t = (-150 - 152.118)/(-64)}}}
You can finish