This was the bonus question on my physics exam. The answer I gave was around 48m. Can some physics fags get the answer right / call me a brainlet?
>You drop a rock in a well, it takes 3 seconds until to hear it. Given that the speed of sound is 343m/s, what is the height of the well?
Jaxson Green
343 m/s * 3 s = 16890 m glad I could hepl
Lincoln Young
but you also have to account for how long the rock takes to fall
Nolan Bell
I simplified the problem by assuming that I knew the correct answer
Jace Foster
why did you answer 48?
shouldnt it just be 343*3=1029
1029 metres
Justin Flores
how long does it take for the rock to fall?
Austin Russell
Does the 3 seconds account for the dropping of the rock
Samuel Perry
Well, I figured, since its in free fall, that it would take some time to reach the bottom, and that maybe, since sound speed is so quick, that the distance covered by the time it takes to reach the bottom is small such that sound reaches your ear quick?
I can give a detailed procedure of what I did, though.
Camden Rogers
faggot OP didnt specify how heavy the rock was. psh go back to school, kid
Joseph Moore
certain elements fall faster or slower than others. its a trick question from the start. i think your teacher is a troll
Austin Robinson
Yes, I assumed this. To clarify further, the time it takes to reach bottom I assume to be less than 3 seconds.
Grayson Cook
Top kek, Do you just work on problems without reading and thinking about what it's asking?
Dylan Jones
y = -0.5g(t)^2 y = 343 (t') t+t' = 3
Now do algebra.
Grayson Rivera
We're using a simplified model where the body is just a point, mass isn't taken into account. What about the experiment on the moon (feather vs cannonball)?
Christopher Hall
I got 40.6 meters, but take this with a grain of salt as I havent done physics in years.
let d be the depth of the well, s be the time the rock spent falling and s be the time the sound spent moving. then d = 1/2 * 9.8 * r^2, d = 343 * s, 3 = s + r. solving gives 40.6 as d.
Bentley Wood
oops r is the time the rock spent falling, not s
Samuel Carter
g = 10 after 2s it's fallen 30 m but 30/343 is much less than 1 after 2.5 s it's fallen 42.5 m 42.5/343 is less than .5 after 2.75 s its fallen 52.5 m 52.5/343 again not .25 after 2.9 s it's fallen 57 m 57/343 is greater than .1 m back to 2.8 s fallen 54 m 54 / 343 < .2 so try 2.85 s....
can someone smarter than me come up with a relationship?
Justin Hill
Ay = g and not -g?
Jace Hill
If I remember correctly, that t was near the t I plugged into y = -.5g(t^2)
Charles Perry
d is the depth of the well, it must be a positive value
Matthew Hernandez
yeah I'm wrong you have to integrate the velocity gt to find the distance
gt^2/2 = c then (3-t)343 = c
another user already provided this
Eli Phillips
so (3-(2c/g)^.5)343=c is right
William Rogers
Fuck, this makes most sense. Lmao I remember doing a quadratic formula and getting a negative time (which yielded that 47m) Thanks user.
Chase Moore
It sounds like you did d = -1/2 ..., which gives d = -48. Really close and frustrating lol. You need either be really, really attentive with your choice of coordinates (which would have caused you to correctly add a negative to the other equation) or rely on good physical intuition. I'm not sure what insight lead me to realize you could ignore signs, let d be the length of the well and figure out how long it would take to fall and echo across that length, but over time you start to intuit these things. I studied physics for six years so I've had time to build up that intuition.
Jose Sullivan
343 / s
Carter Jenkins
>We're using a simplified model where the body is just a point, mass isn't taken into account. then how does sound propagate if nothing has masses
the prof is a retard.
Liam Fisher
There might be a smarter way to do it, but this is how I reasoned it out.
After the rock is dropped, at some time, tf, it reaches the bottom of the well, h. The distance the rock travels should be y=-(g*t^2)/2
So plugging in our known tf and h: h=-(g*tf^2)/2
We know that after the rock reaches the bottom of the well, the sound travels up at 343 m/s, unaffected by gravity and moving at a constant velocity, we know that we can just use d = v*t the distance traveled being h, the velocity being the speed of sound, and t being the time difference between 3 seconds and tf, note that because the rock and the sound travel in opposite directions you need to account for the sign difference, so just through in a negative sign. h = - 343 * (3-tf)
Setting the two equations equal to one another, you have a quadratic, one of the times you get is tf = 2.8812820, plug that back in to the y=-(g*t^2)/2 and you get 40.7202612 m
I think I solved this right, and I hope more work was follow-able.
Ayden Bennett
X-Xo=Xo+Vo*t+at^2/2
use this feggit
Adam Hernandez
Too bad, but you are a brainlet
Michael Baker
Solved it before seeing your posts but yeah, here's the reasoning >pic related
Chase Johnson
>g=10
Nathaniel Reyes
>implies you should ignore air resistance >then gives you the speed of sound in air Pig disgusting, hang your professor.
Blake Ward
yeah and what is the mass of the thing
Levi Smith
first year physics was fun, except for lab lab was chinsey as fuck
Anthony Bailey
> implying engineers use g = 10 m / s^2 instead of g = 32 ft / s^2
Jace Davis
tyson chicken question. the height of the well is exactly 0. level water is the reference
Julian Reyes
...
Jeremiah Walker
I assume you know that [math]D = A/2 * T^2[/math] or if you rearrange the equation [math]T = \sqrt{D/(A/2)}[/math]
so your equation is [math]\sqrt{D/(10/2)} + D/343 = 3[/math] I don't know calculus very well so I can't do this
Jaxson Ramirez
5
Hunter Hall
What programming language is this?
Kayden Barnes
Is air resistance negligible?
Lucas Morales
>g=10 Engineer spotted
Logan Bennett
physicist post doctorate here
easy.
how heavy is the rock brainlet so I can calculate how fast does it fall.
Gavin Johnson
24 mg
Nathan Reed
The derivative of 343m^3 is 1029m^2. 1029 - 343 is 686. The answer is 686
Landon King
the derivative of a number as a function is 0
Jack Richardson
d = .5*g*t^2; t1=sqrt(2d/g) t2=d/v both times add to 3 seconds 3 = sqrt(2d/9.81)+d/(343) 9 -6d/343+d^2/343^2=2d/9.81 9-d(6*9.81+686)/(343*9.81) + d^2/343^2=0 quadratic equation d = 40.71 meters