Physics Problem: Flower Pot Freefall Calculation for Safety

  • Thread starter Thread starter maxwilli06
  • Start date Start date
AI Thread Summary
A physics problem involves calculating how close a flower pot can fall before a warning can reach a man below. The pot falls from a height of 18.4 meters, and the sound of a warning takes 0.05 seconds to reach the man, who requires an additional 0.3 seconds to react. To ensure the man can avoid the pot, the warning must be given 0.35 seconds before impact. Calculations show that the pot will be traveling at approximately 14.9 m/s just before hitting the man, allowing it to fall to about 7.59 meters above the pavement when the warning is given. The discussion highlights the complexity of the problem, with some participants noting discrepancies in calculations.
maxwilli06
Messages
23
Reaction score
0
I was trying to do this physics problem which I thought is really damn
easy but it won't ever count my answer right on webassign i just want to
see what other people get.

A flower pot is knocked off a balcony 18.4 m above the sidewalk and
falls toward an unsuspecting 1.76 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.
 
Physics news on Phys.org
This answer only provided as a reference. I'm no physics ace...
Assuming speed of sound at sea level = 340.3 m/s

There are 17.4m between the balcony and the man, so it takes D/T=R -> 17.4/340.3 = .05s for the sound to reach the (top of the) man. So you need to shout about 0.3s+.05s=0.35s before the pot reaches his head for him to be able to get out of the way in time (time for sound to reach his head and time for him to react.)

The pot will be traveling at Vf^2=Vo^2+2*a*x -> ?^2 = 0+2*9.81*17.14 ->sqrt(336.29) = 18.33m/s. 0.35s before that, the pot will be traveling at 9.81m/s/s*0.35s=3.433m/s slower.

So at the last possible second the pot will be traveling at 14.897m/s. Plugging that into Vf=Vi+A*T we get 14.897=0+9.81*t ->14.897/9.81 = 1.52 seconds. This is the amount of time you can let pass before the pot will inevitable strike the poor man.

So after 1.52 seconds how far will the pot be above the pavement? (14.897^2-0^2)/(2*9.81) = distance from balcony = 11.32 m
Which is 18.9-11.32 = 7.59m above the pavement.

This problem was much harder than it looks. What did you get?
 
Yea, close enough to what I got...7.51m. KBL-8, I think you used the wrong balcony height. It's 16.64m between the balcony and the mans head.
 
Last edited:
Whops. Can't subtract.
 
Thread 'Struggling to make relation between elastic force and height'
Hello guys this is what I tried so far. I used the UTS to calculate the force it needs when the rope tears. My idea was to make a relationship/ function that would give me the force depending on height. Yeah i couldnt find a way to solve it. I also thought about how I could use hooks law (how it was given to me in my script) with the thought of instead of having two part of a rope id have one singular rope from the middle to the top where I could find the difference in height. But the...
Thread 'Voltmeter readings for this circuit with switches'
TL;DR Summary: I would like to know the voltmeter readings on the two resistors separately in the picture in the following cases , When one of the keys is closed When both of them are opened (Knowing that the battery has negligible internal resistance) My thoughts for the first case , one of them must be 12 volt while the other is 0 The second case we'll I think both voltmeter readings should be 12 volt since they are both parallel to the battery and they involve the key within what the...
Back
Top