Question 211479
An object is thrown upward from the top of an 80 foot building with an initial velocity of 64 feet per second. The height h of an object after t seconds is given by the quadratic equation (((h=-16t^2+64t+80))). When will the object hit the ground.


Since we're looking for "h," or the height at 0, we then set the equation to 0, and solve for t. We will then have:

{{{0 = -16t^2 + 64t + 80}}}

{{{-16(0) = -16(t^2 - 4t - 5)}}} ------> factoring out the GCF, -16

{{{0 = t^2 - 4t - 5}}} -------> {{{-16 = -16}}}

0 = (t - 5)(t + 1)


Therefore, t = 5, or -1, but t = - 1 is ignored

This means that the object will hit the ground {{{highlight_green(5)}}} seconds after being thrown in the air.