the air temperature is a nice 20 degrees celsius. If a ball is dropped in a well, the sound is heard 4.0s later, how deep is the well?

Question

This is a Physics 11 word problem. Please send me the full solution. Im really stuck on this.

Answer 1

First, get the speed of sound at 20 C.

 

The simplest formula for the speed of sound in air is

 

v = 20.1 √T (m/s)

 

T = absolute temperature (K) = C + 273 = (20 + 273) K = 293 K 

 

v = 20.1 √293 = 344 m/s

 

y = depth of well

 

Neglecting air friction,

 

y = (1/2) g t1^2 where g = 9.8 m/s^2 and t1 is the time it takes for the ball to reach the bottom of the well

 

In addition,

 

y = v*t2 where v = speed of sound and t2 is the time it takes the sound to travel from the bottom of the well to the top

 

Then t1 + t2 = 4.0 s ⇒ t2 = 4 - t1

 

Therefore,

 

(1/2) g t1^2 = v*t2 = v(4 - t1)

 

(1/2) g t1^2 = v(4 - t1)

 

Now rearrange as a quadratic equation in t1, substitute known values, and solve for t1.

 

(1/2) g t1^2 + v*t1  - 4v = 0

 

(1/2)(9.8) t1^2 + 344*t1 - 4(344) = 0 

 

4.9 t1^2 + 344*t1 - 1376 = 0

 

t1 = {-344 ± √[(344)^2 - 4(4.9)(-1376)]} / (2*4.9)

 

t1 = (1/9.8) (-344 ± 381.1897) s

 

Use the positive square root to get positive time.  Then

 

t1 = 3.7949 s

 

Finally, 

 

y = (1/2) g t1^2 = (1/2) (9.8) (3.7949)^2 s = 70.57 m

 

It should give the same result using y = v*t2

 

y = v*t2 = v*(4 - t1) = 344(4 - 3.7949) m = 70.55 m (slightly different because of roundoff of the square root)

Answer 2

[EDIT:  You should follow Arturo's solution.  It is conceptually identical to mine, and, algebraically, it is a lot more straightforward]

 

There are two processes involved in this action:

 

1.  Dropping the ball into the well;  this is described by kinematics

2.  The sound traveling back out of the well.  We assume this is not described by kinematics, because the sound is not a physical object.  It travels at a constant speed (which we can take to be 343 m/s, since we are told the air temperature is 20^o C

 

[NOTE:  I think Katie's solution is incomplete, as it posits the 4 s as just the time of the ball falling, neglecting the time it takes the sound to travel out of the well; this may seem like a trivial addition, but it makes a significant difference in the calculated depth of the well]

 

Let's look at the second process first.  The second process is described by the equation relating displacement (d), velocity (v), and time (t) for constant-velocity motion.  That is, d = vt.  In this equation, we assume we know v (the speed of sound), leaving d (the depth of the well) and t (the time it takes the sound to get from the bottom of the well to the top) as unknowns.  We can write this mathematically as:

 

[1]  d_well = v_soundt_up

 

where

 

d_well = depth of the well (unknown)

v_sound = speed of sound (known)

t_up = time it takes sound to travel from the bottom to the top of the well (unknown)

 

Since this is only one equation, but with two unknowns, we need more equations relating the unknowns.  Let's look at the first process for this.

 

The first process is a free fall motion, vertical motion under the influence only of gravitational acceleration.  The relationship between displacement and time for this process is given by  d = v_ot - (1/2)gt², where v_o is the initial velocity of the ball.  Since the ball is dropped (released from rest), its initial velocity is 0.  So this kinematic relationship becomes:

 

[2] -d_well = (1/2)(-g)t_down² --> d_well = (1/2)g(t_down²)

 

Here, I have defined downward as negative, and in kinematic equations, direction matters.  So, since the ball is falling, and ends up lower than it starts, d_well is negative in this equation (displacement is down), as well as gravitation acceleration g being negative.

 

Note that we did not quite get, in Equation [2], the second equation we needed relating the same two unknowns as Equation [1].  Because we introduced a third unknown, the time it takes the ball to drop down the well.  So now we need a third equation relating these unknowns to be able to solve for them.

 

Fortunately, we have that third relationship, because we know that the time from the ball's release to when the splash it heard is 4 s.  Thus ,we know:

 

[3]  t_down+t_up = 4 s

 

NOW we can solve for whichever unknown we want (for d_well in this case).  There are several ways this could be done.  I will choose just one of those ways:

 

Start with Equation [3], rearranging it to give t_up as a function of t_down:

 

t_up = 4 - t_down

 

Putting this into Equation [1] gives:

 

[4]  d_well = v_sound(4 - t_down)

 

Then, rearranging Equation [2] to give t_down as a function of d_well gives:

 

t_down = √(2d_well/g)

 

Putting this result into Equation (4) gives:

 

d_well = v_sound(4 - √(2d_well/g))

 

Since v_sound and g are known quantities here, this is now one equation with one unknown (the unknown we want to solve for).  With some algebra, we can solve for d_well.  I arranged the last equation into a quadratic equation and solved with the quadratic formula.  There were two solutions, but only one was not extraneous, and it was

 

d_well = 58.8 m

 

I will check this again when I am more alert :).  But I believe the technique is sound.

 

If you would like to go into more detail or check a solution, I am available for online sessions.  Just let me know!

Answer 3

Hi Meghan,

For this type of problem one can use the equation d=0.5*g*t^2, where g is 9.8 m/s^2, t is the time to fall, and d is the distance fell. Plugging in 4.0s for t, we get d=0.5*(9.8m/s^2)(4.0s)^2. Simplifying, d=0.5*(9.8m/s^s)(16s^2) = 78.4 m. So the well is 78.4 m deep. I should note this is what you do for basic high school physics (where as far as I've experienced, problems are one or two steps and so much physical reality is ignored). I didn't realize physics 11 was apparently higher level. So definitely follow the other guys's suggestions!

Katie