# Homework Help: Determine the actual depth of the shaft to the nearest metre

1. Jan 26, 2005

### kingyof2thejring

Hi there again,

I am having problems answering this question

A girl wishes to estimate the depth d of a straight mine-shaft. She drops a stone vertically down the shaft and finds that there is an interval of 6 seconds between the instant she dropped the stone at the top of the shaft and the time that she heard the stone hit the bottom of the shaft.

(i) She decides to take a first estimate of d by assuming that the stone took 6 seconds to drop from top to bottom of the shaft. Calculate this estimate.

(ii) She then uses her first estimate together with the speed of sound, which is 332 m/s, to estimate the time taken for the stone to drop. Calculate this second estimate of d.

(iii) Determine the actual depth of the shaft to the nearest metre.

s=d
u=0
v=/
a=9.8
t=6

using the following equation
s=ut+1/2at^2
d=1/2 * 9.8 * 6^2
i get d=176.4m
but iam not sure if this is correct though

i am really stuck in this part of the question
here are some random ideas i have tried
t=d/s 176.4/332 = 0.53 s giving me the time taken for the sound to go up
6-0.53= 5.47 s time taken for the stone to drop
s=d
u=0
v=/
a=9.8
t=5.47s

s=ut+1/2at^2
d=1/2 * 9.8 * 5.47^2
d=146.6m

from here i dont know were to go with this question
Any suggestions/ideas will be much appeciated.

2. Jan 26, 2005

### daster

From what I understood from the question, you answers for the first two parts are correct.

For the third part, what are the variables you don't know? Try thinking of the problem as two seperate events.

3. Jan 26, 2005

### Achtung

yep ur first two answers are correct.
as for the third part think of it this way the total time of the stone travelling down the shaft and the sound travelling back up is 6 seconds.
hope this helps.

4. Jan 26, 2005

### daster

Well put.

5. Jan 27, 2005

### kingyof2thejring

thx for sharing your ideas with me

so i should be thinking of two seperate events in this problem.
i thought i had already considered the time of the stone down the shaft and the sound back up being 6 sec in part 2.

i need some more help in understanding what to do with the two seperate events in calculating the actual depth.

6. Jan 27, 2005

### daster

The stone travels d meters down the well in t1 seconds. The sound of the impact travels d meters in t2 seconds. Both events take 6 seconds to happen.

I hope that helps.

7. Jan 27, 2005

### kingyof2thejring

how do i set up simultaneous equations to solve d

8. Jan 27, 2005

### daster

t1+t2=6

When it falls down:
s=ut+0.5t2
d=4.9(t1)2

Can you do the same for the sound?

Last edited by a moderator: Jan 28, 2005
9. Jan 27, 2005

### kingyof2thejring

s=ut + 1/2 a t^2
d=4.9(t1)^2
d=4.9(6-t2)^2

for the sound
-d=-4.9(t2)^2
d=4.9(6-t1)^2

10. Jan 27, 2005

### Achtung

think of it this way:-
let t1 be the time taken for the stone to drop.
let t2 be the time taken for the sound to travell back to the girl.
let d be the length of the mine shaft.

d = 4.9 (t1)^2 (stone falls)
==> t1 = sqrt(d) / sqrt (4.9) --(i)

d = 332 * t2 (sound travels)
==> t2 = d / 332 ---(ii)

t1 + t2 = 6 --(iii)

substitute (i) and (ii) in (iii) and solve for the positive value of the quadratic equation.
in doing so you are solving for sqrt(d) . Square this answer to obtain the value of d.
i got 150.7166298 which to the nearest metre is 151m

11. Jan 27, 2005

### kingyof2thejring

thx
bytheway
did you change
(i) + (ii) = 6 for the quadratic equation
in the form ax^2 + bx + c =0
before getting positive value

12. Jan 27, 2005

### Achtung

yea,
[sqrt (d) / sqrt (4.9) ] + [ d / 332 ] = 6
==>
[(sqrt(d))^2 * sqrt (4.9)] + [332 * sqrt(d)] - [1992 * sqrt (4.9)] = 0

here 'x' = sqrt(d)

13. Jan 27, 2005

### kingyof2thejring

ta
i can dig that