A flower pot is knocked off a window ledge from a height d = 19.6 m above the sidewalk. It falls toward an unsuspecting man of height h = 1.79 m who is standing below. Assume the man requires a time interval of Δt = 0.300 s to respond to the warning. How close to the sidewalk can the flower pot fall before it is too late for a warning shouted from the balcony to reach the man in time?

Question

Physics

Zion Juarez

9 March, 19:16

A flower pot is knocked off a window ledge from a height d = 19.6 m above the sidewalk. It falls toward an unsuspecting man of height h = 1.79 m who is standing below. Assume the man requires a time interval of Δt = 0.300 s to respond to the warning. How close to the sidewalk can the flower pot fall before it is too late for a warning shouted from the balcony to reach the man in time?

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Answers (1)

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Jefferson Cortez · Answer 1

The distance close to the sidewalk can the flower pot fall is

x = 14.83 m

Explanation:

Given

Δt = 0.300 s

d = 19.6 m

h = 1.79 m

Knowing as the velocity of the sound as a 330 m/s

t = (19.6 - 1.79) m / 330 m/s

t = 0.0539 s

Total time

tₙ = 0.3 + 0.0539 = 0.3539 s

Time for flower-pot

s = ¹/₂ * g * t²

tₐ = √[ (2 * 17.81m) / 9.81m/s²]

tₐ = 1.34 s, t' = 0.3539

1.34 - 0.3539 = 0.9861 s

19.6 m - x = ¹/₂ * g * t ²

x = 19.6 - ¹/₂ * (9.81) * (0.9861) ²

x = 14.83 m