# Several phys201 problems (need help on how to approach them)

1. Sep 28, 2005

### bmore007

hey, these problems are for my physics 201 class and i understand what's goin on somewhat i just need help in my approach to the problems. They have to do with one and two dimensional motion and vectors.

#1 A rock is dropped from rest into a well. The sound of the splash is actually heard 2.3 s after the rock is released from rest.(a) How far below the top of the well is the surface of the water? The speed of sound in air (at the ambient temperature) is 336 m/s.(b) If the travel time for the sound is neglected, what percentage error is introduced when the depth of the well is calculated?

#2 solved, thank you

any help would be GREATLY appreciated

Last edited: Sep 29, 2005
2. Sep 28, 2005

### StatusX

1. Find the time it would take you to hear the splash for a well of depth h. This is equal to the time it takes a rock to fall a distance h plue the time it takes sound to travel a distance h. Then solve for h given the time is 2.3 seconds. For the second part, do the same thing ignoring sound travel time and calculate your percentage error.

2. Check to make sure you have the shell falling in the correct direction.

3. Sep 28, 2005

### bmore007

thank you i got #2 correct, i'm still stuck on the first one though

how would i go about finding the time it would take you to hear the splash for a well of depth h?

4. Sep 28, 2005

### hotvette

1a is straightforward. You are given time and asked to find the distance. You should be able to write an equation of motion that relates distance, velocity, and time. Right? You know 2 of the 3 parameters, so solve for distance. Hint: think of the splash sound as a particle that starts moving up when the rock hits the water.

Re 1b, I don't understand it.

Last edited: Sep 28, 2005
5. Sep 29, 2005

### bmore007

"1a is straightforward. You are given time and asked to find the distance. You should be able to write an equation of motion that relates distance, velocity, and time. Right? You know 2 of the 3 parameters, so solve for distance. Hint: think of the splash sound as a particle that starts moving up when the rock hits the water."

here's what i'm doing and so far to no avail

yf= yi + Vot + 1/2(-g)t^2

0 = h + 336(2.3) +1/2(-9.8)(2.3)^2

-h = 772.8 - 25.9

-h = 746.9

am i on the right track or on the wrong train completely?

6. Sep 29, 2005

### hotvette

You are on the right track. What you want is yf - yi. One more thing, sound is massless and isn't acted upon by gravity.

7. Sep 29, 2005

### bmore007

yf-yi = Vot

??

man i'm stupid

8. Sep 29, 2005

### hotvette

You are told both $$v_0$$ and t.

9. Sep 29, 2005

### bmore007

here's what i know

Yi= h
Yf= 0
t= 2.3s
Vo= 336m/s
g = -9.8

and i want to use Yf-Yi= Vot + 1/2(g)^2

correct?

1/2g^2 cancels out because of there being no g

-h = 336(2.3)

i guess my way of using the formula is wrong?

10. Sep 29, 2005

### hotvette

I believe you have it. Wait. I re-read the problem. If I understand correctly, the 2.3 seconds is the time it takes the rock to fall the height of the well into the water + the time for the sound to go from the water to the top of the well. A bit more complicated........

Last edited: Sep 29, 2005
11. Sep 29, 2005

### bmore007

this program is saying my answer is incorrect

12. Sep 29, 2005

### hotvette

Yep, see my edit. Do you know the mass of the rock?

13. Sep 29, 2005

### bmore007

no mass for the rock, everything that is given is in the problem above

14. Sep 29, 2005

### hotvette

I think I have it:

1. get x1 in terms of t1 from the equation of motion, where x1 is the distance the rock travels from the top of the well to the water. Actually, mass doesn't enter into the equation

2. get x2 in terms of t2 from the equation of motion (of the sound), where x2 is the distance that sound travels from the water to the top of the well. In this case, g doesn't enter into the equation

3. set x1 = x2 (i.e. the 2 distances are the same)

4. You know t1 + t2 = 2.3 sec

5. You have 2 equations in 2 unkowns that you can solve for t1 and therefore, for x1

15. Sep 29, 2005

### bmore007

1. get x1 in terms of t1 from the equation of motion, where x1 is the distance the rock travels from the top of the well to the water. Actually, mass doesn't enter into the equation

xf1 = xi1 + Vox1 (t1)

xf1-xi1 = Vox1 (t1)

xf1-xi1/Vox1 = t1

2. get x2 in terms of t2 from the equation of motion (of the sound), where x2 is the distance that sound travels from the water to the top of the well. In this case, g doesn't enter into the equation

xf2 = xi2 + Vox2 (t2)

xf2-xi2 = Vox2 (t2)

xf2-xi2/Vox2 = t2

3. set x1 = x2 (i.e. the 2 distances are the same)

xf1-xi1/Vox1 = xf2-xi2/Vox2

edit: thank you very much, i'm gonna try to pick this back up tomorrow

Last edited: Sep 29, 2005
16. Sep 29, 2005

### hotvette

The falling rock is acted upon by gravity but has no initial velocity:

$$x_1 = \frac{1}{2} g t_1^2$$

The sound particle isn't acted upon by gravity but has an initial velocity:

$$x_2 = v_0 t_2$$

17. Sep 30, 2005

### hotvette

Were you able to solve it? The answer I got for the well depth using the above methodology was somewhere between 20 and 30 m.

Last edited: Sep 30, 2005