Calculating Height of Building w/ Sound Speed

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Homework Help Overview

The discussion revolves around calculating the height of a building based on the time it takes for a rock to fall and the sound of it hitting the ground to be heard. The problem involves concepts from kinematics and sound propagation, specifically relating to free fall and the speed of sound.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the relationship between the total time of 4.8 seconds, the time taken for the rock to fall, and the time for the sound to travel back up. There are attempts to establish equations relating height, time, and speed, with some participants questioning the interpretation of the total time.

Discussion Status

Participants are actively engaging with the problem, offering various equations and methods to relate the variables involved. There is a focus on eliminating variables to isolate the height, with some participants expressing uncertainty about their calculations and seeking clarification on their equations.

Contextual Notes

There is an ongoing discussion about the correct interpretation of the time intervals involved and the need for additional practice in solving similar physics problems. Some participants express confusion over the equations they have derived and their relationships.

Tereza
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Hello,

I'm calculating a physics problem and I'm supposed to calculate the height of a building given that:

rock is dropped from the building,
the sound of the rock hitting the ground is heard 4.8s later,
the speed of sound is 350m/s,

and I don't know how do I calculate the time since the rock hit the ground until it reached my ears.

Thank you!
 
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how do I calculate the time since the rock hit the ground until it reached my ears.
distance over speed.
 
Simon Bridge said:
distance over speed.

I don't have the distance.
 
That's right - distance is what you are trying to find.
You do, however, have the total time, the speed of sound, and the acceleration due to gravity.
 
Is it 4.8 seconds from the time you drop the rock until you hear the sound? Or 4.8 seconds from the time the rock hits the ground until you hear the sound?
 
I'm interpreting that at t=0, rock is dropped, t=T=4.8s hear the sound.
There are four equations and four unknowns ... waiting for OP to realize ;)

Rock accelerates at g=9.8m/s, falls a height h, and hits the ground in time T1 at speed v making a "crack" noise.

From this we see that OP can work out:
v in terms of g and T1 ...that will be eq(1)
h in terms of v and T1 ...that will be eq(2)

The noise returns the same distance at speed c=350m/s in time T2:
height of the building by the speed of sound and T2 ... (3)

The total time between drop and sound is:
T=T1+T2 ...(4)

Four equations and four unknowns.
 
Thank you.

I have : -1/2aT12 = c(4.8-T1)

I still can't find the answer.
 
Well - technically T1 is one of the variables you want to eliminate, you want to keep h because that's what you want to find. I think you may have misplaced a minus sign too.

Please list your four equations... number them.
You will probably instinctively combine the first two... that's fine, you then have three equations and three unknowns.

However: notice that you know a and c, so you can get T1 from that equation. Substitute that number into all your equations where T1 appears and you now have three equations and three unknowns.

But it is better to to use the first three equations to eliminate all the unknowns except for h. Have you solved simultaneous equations before?
 
Last edited:
You give me wonderful ideas on how to start solving physics problems effectively, but I need so much more practice.

This time I got: T2= 2h/a + h2/c2

however, I still can't get it. It's always so close.

I don't know what more to ask you, I think you've told me all that you could.

Thank you!
 
  • #10
That does not look right to me - I need to see your working or I cannot help you :(

I suspect you actually had $$T=\sqrt{\frac{2h}{g}}+\frac{h}{c}$$ and you squared both sides. Is that the case?
If so then that was a mistake.

But look at the form of your final equation ... you've seen it before.
What do you call it when the variable is squared in a function: what sort of function is it?
It has form ##Ax^2+Bx+C=0##
 
Last edited:
  • #11
I think I have it now.

1/2at12=ct2

and t1= t-t2

Then solve quadratic equation with a=1; b= -71.4; c=23.04

Thank you so much!
 
  • #12
No worries.
 

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