Question 204099
{{{ S(t)= -16t^2-32t+128}}}
Looking at the equation, when {{{t = 0}}},
{{{S(t) = 128}}} as it should
When {{{S(t) = 0}}}, that is when the 
wrench hits the ground
I can find {{{t}}} by
{{{0 = -16t^2 - 32t + 128}}}
Divide both sides by {{{16}}}
{{{-t^2 - 2t + 8 = 0}}}
Use the qudratic equation
{{{t = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
{{{a = -1}}}
{{{b = -2}}}
{{{c = 8}}}
{{{t = (-(-2) +- sqrt( (-2)^2-4*(-1)*8 ))/(2*(-1)) }}}
{{{t = (2 +- sqrt( 4 + 32 ))/(-2)) }}}
{{{t = (2 +- 6)/(-2)}}}
{{{t = -4}}} (not possible)
{{{t = 2}}} answer
It takes 2 sec for the wrench to hit the ground
after {{{1}}} sec,
{{{ S(t)= -16t^2-32t+128}}}
{{{ S(1)= -16*1^2-32*1+128}}}
{{{S(1) = -16 - 32 + 128}}}
{{{S(1) = 80}}}
After 1 sec, the wrench is 80 ft above the ground