# This was the bonus question on my physics exam

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?

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343 m/s * 3 s = 16890 m

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

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

shouldnt it just be 343*3=1029

1029 metres

how long does it take for the rock to fall?

Does the 3 seconds account for the dropping of the rock

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.

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

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

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

y = -0.5g(t)^2
y = 343 (t')
t+t' = 3

Now do algebra.

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)?

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.

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

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?

Ay = g and not -g?

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

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

yeah I'm wrong you have to integrate the velocity gt to find the distance

gt^2/2 = c
then (3-t)343 = c

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

Fuck, this makes most sense.
Lmao I remember doing a quadratic formula and getting a negative time (which yielded that 47m)
Thanks user.

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.

343 / s

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.

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.

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

use this feggit

Too bad, but you are a brainlet

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

g=10

implies you should ignore air resistance
then gives you the speed of sound in air

yeah and what is the mass of the thing

first year physics was fun, except for lab
lab was chinsey as fuck

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

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

I assume you know that $D = A/2 * T^2$
or if you rearrange the equation $T = \sqrt{D/(A/2)}$

so your equation is $\sqrt{D/(10/2)} + D/343 = 3$
I don't know calculus very well so I can't do this

5

What programming language is this?

Is air resistance negligible?

g=10
Engineer spotted

physicist post doctorate here

easy.

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

24 mg

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

the derivative of a number as a function is 0

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
d = 40.71 meters