Physics Question?

Question

A rock is dropped into a well from rest, after 2.4 seconds the sound of the splash can be heard, what the the depth of the well?

the time isnt "the time it takes to get to the bottom" the time is " time it takes to get to the bottom+time it takes for the splash to be heard"

you cant just multiply everythign together.....

Answer 1

In have considered the time needed for sound to travel back the distance.

Suppose:

X= depth of well
T1= time needed for the stone to get to the end of well
T2= time needed for sound to get back the same distance

The velocity of sound is like 430m/s^2


1- For stone:
X=(1/2)*9.8*(T1)^2

2- For Sound:
X=340*T2

3- We know:
T1+T2=2.4

Solving the above 3 equations:

X=4.9T1^2
X=340(2.4-T1)

So:
4.9T1^2=-340*T1+816

The other arrangement:
4.9T1^2+340T1-816=0

T1= 2.32s

Now we use eq. (1):

X=4.9(2.32)^2= 26.43m

Depth of well is 26.43m

Answer 2

well

you first have the time it takes for the stone to fall. you can use the equations of accelerated movement, such as x = (1/2)at^2, which will give you t1=sqrt(2x/a). Where x is the depth of the well obviously.

then the time it takes for the sound of the splash to come to your ears, which is is simply the distance divided by the speed of sound, which i'll call "s".
t2=x/s

so the total time, t, is t1+t2

so t=sqrt(2x/a)+(x/s)

you know t, and you can find the speed of sound, so this leaves you with a quadratic equation of the form ax^2 + bx +c = 0

which you can solve using the famous formula

x=(-b +/- sqrt(b^2 - 4ac))/2a

sorry i'm a bit lazy to do it but it's pertty straightforward

good luck

a

Answer 3

y(t) = y0 + v*t + .5*a*t^2, right?

the y0 is 0 and rock is dropped from rest so v=0, thus...

y(t)=.5*a*t^2

a is acceleration due to gravity in this case... -9.8m/s^2. and t is 2.4s, so...

y(2.4) = .5*-9.8*(2.4)^2
(plug that into a calculator, I'm not going to do that for you)

you can ignore the negative sign... it only refers to direction but the question asks for a distance.

*edit* even though they say that, they probably mean to assume the time for the sound to come back is negligible... if you don't make this assumption then you would have to make an assumption as to the elevation of the well anyway(as the speed of sound varies at different pressures).

Answer 4

be very carefull. one has tried to tell you but the other got it all wrong. first of all you will not find the depth of the well but the distance from the its edge to the surface of the water.
now t = t1+t2.
t1 is the time that it takes for the rock to reach the water surface and t2 is the time that it takes sound to travel back to youin order to hear the splash..idelay etc etc etc no friction etc. the sound travles at 342m/sec assumoing constant speed. directly you get that H (well depth") = 342 * t2
but H = 1/2 g (t2)^2
t=t1+t2 = 2.4
and t1=t-t2 or
t2=t-t1 (it is up to you how you will do the maths)

i done it my self. please do it by your own and find the answer. the whole trick though in physics is to realize the concept, no tthe algebra. (although i is very important in order to "find your way through" the equation and work out results)!
i hope that this helps

Answer 5

The equation you need is
s = (1/2)gt² where s is distance, g is the acceleration (in this case, gravity, at 9.8 m/s²) and t is time in seconds.

This gives (1/2)*9.8*(2.4)² = 28.224 meters


Doug

Answer 6

distance formula:

s = 1/2 g t^2

where g is the gravitational acceleration taken as 9.8 ms^-2

s = 1/2 (9.8) (2.4)^2

s = 28.224 m

Answer 7

The distance is equal to: 1/2 g*t^2
g = 9.81 m/s^2
t = 2.4s

28.2528 meters

Answer 8

easy
use Newton's law
i don't recall the formula but
use it with 9.81 as your gravity