Question

The height of a small rock falling from the top of a 124-ft-tall building with an initial downward velocity of –30 ft/sec is modeled by the equation h(t) = –16t2 – 30t + 124, where t is the time in seconds. For which interval of time does the rock remain in the air?
t = 2
–2 < t < 0
0 < t < 2
t > 2

Answer 1

Set h(t)=0: -16t^2 - 30t + 124 = 0 → 8t^2 + 15t - 62 = 0.
Solve: t = [-15 ± √2209]/16 = (-15 ± 47)/16, giving t = 2 and t = -31/8.
Physical time is t ≥ 0, so the rock is in the air for 0 < t < 2 seconds.