Grade 11 Physic Help?

Question

if a rock is dropped down a well, and is heard 16 seconds later, how deep is the well in metres? The speed of sound is 345 m/s. Could someone plead explain to me how to figure this out (full solution) Thank you very much!

Answer 1

Okay I think the other 2 answers you already had when I started this misunderstood your question so I'll get it right for you.

Two things happen in the 16 seconds.
1. rock falls s meters
2. sound travels back up s meters from water surface to well opening

First let's deal with the #1 part, the rock's fall. For this let's use a variable called t1 for time.

We need to use this formula: s = (v1)(t) + ½(a)(t)²
s = displacement, v1 = initial velocity, t=time, a=acceleration

Since the rock is dropped it has no initial velocity. Let down be positive.

s = (v1)(t1) + ½(a)(t1)²
s = (0)(t1) + ½(a)(t1)²
s = ½(a)(t1)²

Okay so we figured out what we can in this step about t1. Now let's go to step 2 and see what we can figure out about t2.

The sound travelled from the water up to the well opening (s meters) in some amount of time we don't know yet (t2). We can describe the sound's movement with this formula:

s = (v)(t) (we use this because the sound is not accelerating)

s = displacement, v = velocity, t = time

Now let's use our variables (t2, our s) and use v2 for the sound's velocity and see what we can put together with the sound.

s = (v2)(t2)

Okay so we know something about the sound and rock now, let's gather what we found and look at it.

1. s = ½(a)(t1)²
2. s = (v2)(t2)

There is something else we know too and can write down as an equation. We know that the time it took for the rock to hit the water and the time it took for the sound of the splash to travel back up to the opening add up to a total of 16 seconds. So we can write it as:

t1 + t2 = 16s (let's isolate the t2 and call it equation 3)
3. t2 = 16 - t1

So all together we have:

1. s = ½(a)(t1)²
2. s = (v2)(t2)
3. t2 = 16 - t1

We have enough equations now to solve for the number of variables we have. Start by substituting t2 from #3 into #2 to get equation #4.

4. s = (v2)(16 - t1)

Now set 1. and 4. equal to each other (because they are both s so they are equal so we can do that)

(v2)(16 - t1) = ½(a)(t1)² (expand it out now)

16(v2)-(t1)(v2) = ½(a)(t1²)

Now put in what we know (v2 = 345m/s, a = 9.8m/s)

16(345)-(t1)(345) = 4.9(t1²) (move it all to one side and it should look familiar)

4.9t1²+345t1-5520=0 (it's a quadratic equation and we can solve those, stick it in your calculator)

t1 solves to
t1 = 13.436
or
t1 = -83.844 (forget this one, the rock didn't hit before you dropped it)

So t1 = 13.436m (we only know this to 2 significant digits)

Now we can finally solve for s by using equation #1
1. s = ½(a)(t1)²
s = ½(9.8)(13.436)²
s = 884.57m (only to two sig. digits)

s = 880 meters

Answer 2

I suppose the time given is from the instant the rock is dropped. If the rock takes t sec to reach the water surface, the depth of the water surface s = 1/2 g*t^2.
Since the total time to hear the sound is 16 sec, the sound takes (16 - t)sec to cover the distance s. That is s = 345*(16 - t).
From these two equations,
1/2 g*t^2 = 345*(16-t), we get
t = 13.4 s. That means the sound wave took 2.6 sec to reach the observer. Therefore
s = 345*2.6 = 897 m.
A very deep well indeed. I am surprised that any sound was heard at all!

Answer 3

I won't give you answer, but I WILL walk you through how to get it

The trick here is in realizing that you have three different equations with a total of three unknowns. The obvious one is the well depth. The other two are the time it spends falling and the time it takes the sound to travel back up. Once you realize that you can write down 3 equations:



t_falling+t_sound=16 (total time)

This first equation is simple, but is also the key guiding insight to the whole problem. It seems like you are making it more complicated because you are taking one value (16 seconds) and splitting it into two variables. Once you write it down, you realize you simply need equations for the falling time and the sound travel time.

1/2g*t_falling^2=Distance
t_sound=Distance/345

Now you have 3 equations and three unknowns so you can solve for "distance". You can do it the fancy pants way of adding equations, but simply substituting one variable out for an expression of itself will do it more quickly. For instance t_sound= 16- t_falling, so you can put that in the third equation and get

16-t_falling=Distance/345

Then you can find an expression for t_falling and plug that into the

1/2g*t_falling^2=distance

equation.

So, work the math, and when you see this kind of problem in the future, ask yourself what equations you can write down that describe what is going on.

Answer 4

Hi There

The formula you need to use is D=ST where D is the distance fallen in meters, S is the speed in meters per second and T is the time in seconds.
So If T=16s and S=345m/s
D= 16s X 345m/s = 5520 m. You can re arrange the formula depending on what parameter you need to determine.

As a quick check to see if you have the formula correct is to look at the units of measurment ie m/s, s, m etc.

In the above case we are looking for the depth of the well ie meters (m)
So agian
D=ST from this we get D (m)=S (m/s) X T (s).
From this the seconds (s) cancel each other out and we are left with meters (m).

Hope this helps and good luck. Physics is such a great subject just keep going

Answer 5

Both the first two answers are wrong. It takes time for the rock to hit the water and then time for the sound to reach the top of the well. It's a two part problem. You have to set them equal to one another because the height falling has to equal the height of sound comming back up. The times will be different so you have 2 values for t so
t1 +t2=16
1/2at1^2+vt1=345(t2)
substitute for t2
1/2at1^2 +vt1=345(16-t1)

now just do the math

Answer 6

Interesting question. There are two parts to the answer. How long did it take for the rock to hit the bottom? And how long did it take for the sound to come back?

First T1= time travelling down, T2= time travelling up.

So T1 + T2 = 16 or T2 = 16-T1

The distance a falling object travels is

D=(1/2)AT1^2 or D=(1/2)(9.8m/s)(T1^2) or
D=4.9m/s(T1^2)

The distance for the sound to come back up the well is

345m/s(T2)

since the distance down equals the distance up

4.9m/s(T1^2) = 345m/s(T2)

Simplify a bit 4.9T1^2 = 345T2 then

4.9T1^2 = 345(16-T1) or 4.9T1^2 = 5520-345T1

4.9T1^2 + 345T1 - 5520 = 0

T1^2 + 70.4082T1 - 1126.53 = 0

T1 = 13.436 sec, T2 = 2.564 sec

So distance = 2.564sec(345m/sec) = 884.578 meters

Answer 7

I had physics in 9th grade and we learned the same thing. When you take physics for a second year usually they take a little from last year and build on it, like moving from one unit to the other. They will probably stray away from optics next year for you.

Answer 8

the velocity of anything in the world is
v=S/t

s is the distance cut my the object
t is the time 2 cut that distance

so the solution will be

s=345/16 = 21.56 m

Answer 9

You have to find the speed of sound in water. Is that the speed they gave?
then use the speed = distance * time

Answer 10

The well must be dry, otherwise we'd need to know when it hits bottom as well as when it hits water.