You throw a rock off the top of a building with an initial velocity of 10.0 m/s. It takes 12.0 s to hit the ground. How tall is the building?

Answer 1

h(t) = -4.9t^2 +vot + ho where vo = initial velocity, ho = initial height, t= time in seconds

h(12) = -4.9(12)^2 + 12(10) + ho = 0

-705.6 + 120 + ho = 0

ho = 705.6 - 120 = 585.6 meters high

UNLESS, the rock was thrown down, not up, then

h(12) = -705.6 -120 + ho = 0

ho = 825.6 meters high

the problem is a little ambiguous. Normally if you threw a rock off a tall building, you wouldn't be throwing it vertically upward. that seems weird. Downward makes more sense.

Answer 2

y(t) = h + v₀t - gt²/2 = h + 10t - 4.9t²; y(12) = h + 120 - 4.9·144 = 0; h = 585.6 m