Question 1058228
y=0.  Note, the square term is always negative.
-16t^2+176t=0
-16t(t-11)=0
t=0, t=11
t=0 is the starting point; t=11 the end point, when it hits the ground.
When the projectile is 16 feet off the ground, -16t^2+176t=16
-16t^2+176t-16=0
factor out a -16
-16(t^2-11t+1)=0
t=(1/2)(11+/- sqrt (121-4); sqrt (117)=10.82
t=(1/2)(11+/- 10.82)
t=(1/2)(21.82) or (1/2)(0.18)
t=0.09 sec and 10.91 sec (note symmetry)
{{{graph(300,300,-10,12,-20,500,-16x^2+176x-16)}}}
{{{graph(300,300,-2,2,-20,500,-16x^2+176x-16)}}}