A brick is thrown vertically upward with an initial speed of 3.00 m/s from the roof of a building. If the building is 78.4 m tall, how much time passes before the brick lands on the ground?

0.918 sec

Explanation:

time of flight:

t=u²/g

=3²/9.8=0.918sec

4.32 seconds

Explanation:

Given:

y = 0 m

y₀ = 78.4 m

v₀ = 3.00 m/s

a = - 9.8 m/s²

Find: t

y = y₀ + v₀ t + ½ at²

0 = 78.4 + 3.00 t - 4.9 t²

4.9t² - 3t - 78.4 = 0

Solve with quadratic formula:

t = [ 3 ± √ (9 - 4 (4.9) (-78.4)) ] / 9.8

t ≈ - 3.71, 4.32

Since t must be positive, t = 4.32. So it takes 4.32 seconds for the brick to land.