Question 17089
The height, h, of an object thrown upward from an initial height, Ho, with an initial velocity, Vo, is given by the function h as a function of time, t:

{{{h(t) = -16t^2 + Vot + Ho}}}

Since Vo = 48 ft/sec and Ho 64 ft, then:

{{{h(t) = -16t^2 + 48t + 64}}} You want to know at what time, t, will the rock hit the ground (h = 0). Set the above function = 0.

{{{-16t^2 + 48t + 64 = 0}}} Solve this quadratic equation for t. First, factor -16.
{{{-16(t^2 - 3t - 4) = 0}}} Apply the zero products principle.
{{{t^2 - 3t - 4 = 0}}} Factor.
{{{(t - 4)(t + 1)}}} Again, apply the zero products principle.
{{{t - 4 = 0}}} and/or {{{t + 1 = 0}}}

If t - 4 = 0, then t = 4 seconds
If t + 1 = 0, then t = -1 second...Discard this solution as negative time is not meaningful in this problem.

The rock hits the ground after 4 seconds.