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?
Accepted Solution
A:
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 ---