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?

9 days ago

343 m/s * 3 s = 16890 m

glad I could hepl

9 days ago

@Boy_vs_Girl

but you also have to account for how long the rock takes to fall

9 days ago

@MPmaster

I simplified the problem by assuming that I knew the correct answer

9 days ago

@Lord_Tryzalot

why did you answer 48?

shouldnt it just be 343*3=1029

1029 metres

9 days ago

@MPmaster

how long does it take for the rock to fall?

9 days ago

Does the 3 seconds account for the dropping of the rock

9 days ago

@Spamalot

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.

9 days ago

>>9228408

faggot OP didnt specify how heavy the rock was. psh go back to school, kid

9 days ago

@viagrandad

certain elements fall faster or slower than others. its a trick question from the start. i think your teacher is a troll

9 days ago

@Evil_kitten

Yes, I assumed this.

@viagrandad

To clarify further, the time it takes to reach bottom I assume to be less than 3 seconds.

9 days ago

@happy_sad

Top kek, Do you just work on problems without reading and thinking about what it's asking?

9 days ago

y = -0.5g(t)^2

y = 343 (t')

t+t' = 3

Now do algebra.

9 days ago

@happy_sad

We're using a simplified model where the body is just a point, mass isn't taken into account.

@SomethingNew

What about the experiment on the moon (feather vs cannonball)?

9 days ago

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.

9 days ago

@Bidwell

oops r is the time the rock spent falling, not s

9 days ago

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?

9 days ago

@Bidwell

Ay = g and not -g?

9 days ago

@likme

If I remember correctly, that t was near the t I plugged into y = -.5g(t^2)

9 days ago

@kizzmybutt

d is the depth of the well, it must be a positive value

9 days ago

@likme

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

9 days ago

@Techpill

so (3-(2c/g)^.5)343=c

@Bidwell

is right

9 days ago

@Bidwell

Fuck, this makes most sense.

Lmao I remember doing a quadratic formula and getting a negative time (which yielded that 47m)

Thanks user.

9 days ago

@Carnalpleasure

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.

9 days ago

@w8t4u

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.

9 days ago

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.

9 days ago

@Lord_Tryzalot

X-Xo=Xo+Vo*t+at^2/2

use this feggit

9 days ago

@CouchChiller

Too bad, but you are a brainlet

9 days ago

@idontknow

@Techpill

Solved it before seeing your posts but yeah, here's the reasoning

pic related

9 days ago

@Lord_Tryzalot

implies you should ignore air resistance

then gives you the speed of sound in air

Pig disgusting, hang your professor.

9 days ago

@idontknow

yeah and what is the mass of the thing

9 days ago

@Lord_Tryzalot

first year physics was fun, except for lab

lab was chinsey as fuck

9 days ago

@Emberfire

implying engineers use g = 10 m / s^2 instead of g = 32 ft / s^2

9 days ago

tyson chicken question. the height of the well is exactly 0. level water is the reference

9 days ago

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

9 days ago

@Ignoramus

What programming language is this?

9 days ago

@Lord_Tryzalot

Is air resistance negligible?

9 days ago

@likme

g=10

Engineer spotted

9 days ago

@Lord_Tryzalot

physicist post doctorate here

easy.

how heavy is the rock brainlet so I can calculate how fast does it fall.

9 days ago

The derivative of 343m^3 is 1029m^2. 1029 - 343 is 686. The answer is 686

9 days ago

@Stupidasole

the derivative of a number as a function is 0

8 days ago

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

8 days ago