Question 892229
{{{h=-16t^2 + vt + s}}} with
{{{t}}}=time in seconds,
{{{h}}}= height in feet above the ground,
{{{s}}}= initial height in feet above the ground (at time zero), and
{{{v}}}= initial velocityin feet per second,

In this case {{{v=2}}} and {{{s=0}}} , so
{{{h=-16t^2 + 2t + 0}}} or simply {{{h=-16t^2 + 2t}}}
Factoring, we get
{{{h=(-16t + 2)*t}}}
When the frog is on the ground, {{{h=0}}} , and
{{{0=(-16t + 2)*t}}} has 2 solutions: {{{system(t=0,"or",-16t+2=0)}}}
{{{t=0}}} is the time when the frog pushed off the gound.
The solution to {{{-16t+2=0}}} gives us the time the frog is on the ground again.
{{{-16t+2=0}}}-->{{{2=16t}}}-->{{{2/16=t}}}-->{{{highlight(t=1/8)}}} or {{{highlight(t=0.125)}}}
The frog lands after {{{1/8}}} of a second (or 0.125 seconds).