Calculating Height of Building w/ Sound Speed

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To calculate the height of a building using the time it takes for a rock to fall and the sound to travel back, the total time of 4.8 seconds includes both the fall time (T1) and the sound travel time (T2). The equations of motion under gravity and the speed of sound are used to derive the relationships between these variables. By eliminating T1, the problem can be simplified to focus on the height (h) of the building. A quadratic equation arises from the calculations, allowing for the height to be solved. The discussion emphasizes the importance of correctly applying physics equations to arrive at the solution.
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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?
 
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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##
 
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  • #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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