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 tal
l, how much time passes before the brick lands on the ground?
2 answers:
Answer:
0.918 sec
Explanation:
time of flight:
t=u²/g
=3²/9.8=0.918sec
Answer:
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.
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