The height of a rock thrown off a cliff can be modeled by h=-16t^2-8t+120, where h is the height in feet and t is time in seconds. How long does it take the rock to reach the ground?

The rock hits the ground after 4 seconds.----------------------Here is the work in how you can find this answer...:----------------------The height, h, of an object thrown upward from an initial height, H, with an initial velocity, Vo, is given by the function h as a function of time, t: h(t) = -16t^2 + Vot + H Since Vo = 48 ft/sec and H 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. ---OVERALL = THE ROCK HITS THE GROUND AFTER 4 SECONDS---