Question 71246
After t seconds, the height of an object with an initial upward velocity of {{{v[o]}}} feet per second and an initial height of {{{h[o]}}} feet is given by the function: {{{s(t) = -16t^2+v[o]t+h[o]}}}
In your problem, {{{v[o] = 64}}}ft/sec and {{{h[o] = 80}}}ft, so you have:
{{{s(t) = -16t^2+64t+80}}}
You want to know at what time, t, will the height, h, be 0, so set the function s(t) = 0 and solve for t.
{{{-16t^2+64t+80 = 0}}} You can simplify this a bit by factoring out -16.
{{{-16(t^2-4t-5) = 0}}} so:
{{{t^2-4t-5 = 0}}} You can factor this quadratic equation.
{{{(t+1)(t-5) = 0}}} Apply the zero product principle.
{{{t+1 = 0}}} and/or {{{t-5 = 0}}}
If {{{t+1 = 0}}} then {{{t = -1}}} Discard this solution as the time must be a positive value.
If {{{t-5 = 0}}} then {{{t = 5}}}
So the object will reach ground-level (h=0) in 5 seconds.