Can Sound Travel Fast Enough to Warn a Man of a Falling Flower Pot?

AI Thread Summary
A flower pot falling from a height of 21.6 meters poses a risk to a man standing below, who requires 0.300 seconds to respond to a warning shout. Calculations show that the total time for the pot to fall must account for both the sound travel time and the man's reaction time. The correct approach involves determining the time it takes for the pot to reach the man's head and adjusting the fall distance accordingly. The original calculations were flawed due to incorrect values being used, which led to inaccurate results. With proper adjustments and algebraic expressions, the issues were resolved, leading to correct answers for both the falling pot scenario and the sound wave displacement amplitude.
buttterfly41
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A flower pot is knocked off a balcony 21.6 m above the sidewalk and falls toward an unsuspecting 1.79 m-tall man who is standing below. How close to the sidewalk can the flower pot fall before it is too late for a shouted warning from the balcony to reach the man in time? Assume that the man below requires 0.300 s to respond to the warning.

So i (21.6-1.79)m / 343m/s = .05776s + .3s = .35776s to yell and have the man react
then, 21.6m= 1/2 * 9.8m/s2 *T^2 ... T= 2.0996 - .35776s = 1.74181s for latest time to wait before yelling,

so i thought the answer should be: 1/2 * 9.8m/s2 * 1.7418^2 = 14.86m down, so 5.74m from the ground... but that is not correct :(

any ideas where i went wrong?

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Also, A sound wave in air has a pressure amplitude equal to 3.94 X10^-3 Pa. Calculate the displacement amplitude of the wave at a frequency of 10.5 kHz.

(Note: Use the following values, as needed. The equilibrium density of air is p= 1.20 kg/m3; the speed of sound in air is v = 343 m/s. Pressure variations P are measured relative to atmospheric pressure, 1.013 x10^5 Pa.)



so, i thought i would use the equation deltaPmax= pwvsmax
and then i plugged the numbers in (w=2pif=65973)... 1.013E5 = 1.2*65973*343*smax
smax= .00373m ... but again, wrong... so i don't know where to go from here

any help would be greatly appreciated. Thanks
 
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buttterfly41 said:
A flower pot is knocked off a balcony 21.6 m above the sidewalk and falls toward an unsuspecting 1.79 m-tall man who is standing below. How close to the sidewalk can the flower pot fall before it is too late for a shouted warning from the balcony to reach the man in time? Assume that the man below requires 0.300 s to respond to the warning.

So i (21.6-1.79)m / 343m/s = .05776s + .3s = .35776s to yell and have the man react
then, 21.6m= 1/2 * 9.8m/s2 *T^2 ... T= 2.0996 - .35776s = 1.74181s for latest time to wait before yelling,

so i thought the answer should be: 1/2 * 9.8m/s2 * 1.7418^2 = 14.86m down, so 5.74m from the ground... but that is not correct.
In physics one has to do the analysis first using algebra. Once you have the physics down, plug in numbers but not before. It makes it easier to see and it usually saves a lot of work.

First of all, as you have correctly noted, the time required is the time it takes for the sound to travel from the balcony to the man + the response time of the man. What is the total time of fall before the pot reaches the position of the man's head? So what is the time that the pot can fall before the warning is given? Write out the algebraic expression for that time and plug in the numbers. [Hint: the total distance of fall is not 21.6 metres. I think this is where you went wrong]

Second, relate the time it has been falling to the distance it has fallen. Write the algebraic expression for that distance. Plug in your numbers.

AM
 
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thank you very much... i fixed both of my problems with your adivce... because for both of them i put in one wrong number for a coefficient... so again, thank you for you help ;0

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