alg. 2 hw help??

Question

an object is thrown upward with an initial velocity of 80 feet per second . how long is the object above 96 ft? use the formula s= -16t(squared) + 80t

Answer 1

For this kind of a problem, you have to assume that the object follows a perfect parabolic curve - no wind resistance or other interference, and the object ends up at the same vertical position as it started (i.e. if you throw it up starting 5 feet off the ground, it ends 5 feet off the ground).

As for finding out when it got to 96 feet, we can plug in 96 for s and solve for the time t:

96 = -16t^2 + 80t
0 = 16t^2 - 80t + 96
0 = t^2 - 5t + 6
0 = (t - 2)(t - 3)
t = 2, 3

So the object is at 96 feet after 2 seconds (on its way up) and again at 3 seconds (on its way down). The total time that it is above 96 feet, then, is 1 second.