A flower pot is knocked off a window ledge from a height d = 19.6 m above the si

Question

A flower pot is knocked off a window ledge from a height d = 19.6 m above the sidewalk as shown in the figure below. It falls toward an unsuspecting man of height h = 1.80 m who is standing below. Assume the man requires a time interval of At = 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? Try approaching the problem in terms of the time at which the shout must occur and then use this to determine the distance, m

Explanation / Answer

Let the Velocity of the Sound = 343 m/sec Here Distance travelled by the Sound to reach the ears of the ment = 19.6 - 1.80 = 17.8 m Therefore Time taken by the Sound to reach to the Ears of the person = Distance/Velocity = 17.8/343 = 0.0519 sec Reaction time = 0.300 sec So Total Time taken by the man to respond = 0.300 + 0.0519 = 0.3519 sec Let Time taken by the port to reach at the level of the Head of the person = t sec Therefore S = ut + 0.5*g*t^2 Initial velocity of the pot, u = 0 m/sec Therefore 17.8 = 0.5*9.8*t^2 t = 1.906 sec Therefore Time Difference = 1.906 - 0.3519 = 1.554 sec So Distance travelled by the pot in this time span = 0.5*9.8*1.554^2 = 11.833 m Required Distance = 19.6 - 11.833 = 7.767 m

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