if you spit down a a really deep hole and you hear it five seconds later. how far down would that be?

Question

i saw some people using this method a few times but never really knew how they calculated it.
if you released an object without throwing it, 10 seconds later it would reach 98 miles per hour.

heres an other question if its more easily understandable.
You drop a stone into a deep well and hear it hit the bottom 5 s later. This is the time it takes for the stone to fall to the bottom of the well, plus the time it takes for the sound of the stone hitting the bottom to reach you. Sound travels about 343 m/s in air. How deep is the well?

Answer 1

fall: h=4.9t^2

hear: h=343(5-t)

h=h

4.9t^2=343(5-t)

4.9t^2+343t-1715=0

t^2+70*t-350=0



t= 4.6863[s] (fall)

t= 5-4.6863= 0.3137[s] (hear)



h= 4.9t^2= 4.9*4.6863^2= 107.6[m]

h= 343*0.3137= 107.6[m]

Answer 2

The hole is about 122m deep (409ft). Sound can be treated as negligible. If you were to include sound and friction the hole would probably be only just over 100m.

They calculate this by either using calculus, or SUVAT equations (simplified calculus)

The SUVAT Equations required for this are

V = U + AT

V^2 = U^2 + 2AS

Where V is final speed, U is initial speed, A is accelleration (usually 9.81 from gravity), S is distance and T is time.



98 Miles per hour isn't quite correct, I say this because it sounds suspiciously like 98 metres per second which you would get from entering 9.81 and 10 into V=U+AT. Without friction, this would allow you to reach speeds of 350kmph

Answer 3

To solve the first problem use:-

v = u +gt

Where u = zero m/s because the object is released and no thrown down with any velocity.

v = 0 +9.81 x 10

v = 98.1 m/s

1 cm = 2.54 cm

1 foot = 12 x 1cm = 12 x 2.54 cm

1yd = 3 x 12 x 1cm = 3 x 12 x 2.54cm

1 mile = 1760 x 3 x 12 x 1cm = 1760 x 3 x 12 x 2.54 cm

1 mile = 160934.4 cm

1 mile = 1609.344 m

Therefore, 1 m = 1/1609.344 miles

Therefore 98.1 m/s = 98.1/1609.344 miles/sec

Therefore 98.1 m/s = 98.1 x 60 m/min = (60 x 98.1)/1609.344 mines /min

Therefore 98.1 m/s= 98.1 x 60 x 60 m/hour = (60 x 60 x 98.1)/1609.344 miles / hour

98.1 m/s= (60 x 60 x 98.1)/1609.344 miles / hour

98.1 m/s= 219.44 miles per hour.



To solve the second problem you need to be able to solve a quadratic.



Let the time taken for the stone to reach the water be tw and let the time taken for splash to reach the ears be ts. Now let T be sum of these two so that

T =tw+ts. Where T is your 5 seconds.



For the stone to reach the water we use:-

s=ut+0.5g(tw)^2

But, since u, the initial velocity of the stone is zero then this reduces to:-

s=0.5gtw^2



For the sound to reach the ears from the water's surface we use :- s =vs x ts



Both of these 's' are the same so we can equate the two expressions giving:-

0.5gtw^2=vs x ts

0.5gtw^2-vs x ts =0

Using T =tw+ts and substituting for ts gives:-

0.5gtw^2-vs (T-tw)=0

0.5gtw^2-vsT +vstw =0

0.5gtw^2 +vstw-vsT=0

0.5gtw^2+343tw -343x5=0

0.5xgxtw^2 +343tw-1715 =0

This is a quadratic equ where:-

a=4.905; b=343 and c=-1715.



You plug these figures into the quadratic formula to give you twp values for tw. One of these values will be less than 5 seconds but greater than zero and this is the correct one of the two. Next, you use this value of tw in

s=0.5gtw^2 to give you the value for s: how deep the well is.

(-b+/ root [b^2 -4ac])/2a

(-343 +/- root[343^2 - (4 x 4.905 x -1715)])/2 x 4.905

(-343 +/- root[117,649 + 33,646])/9.81

(-343 +/- root[151,295])/9.81

(-343 +/- 388.97)/9.81

two solutions:-

(-343 + 388.97)/9.81 = 45.97/9.81= 4.686 sec

or,

(-343 - 388.97)/9.81 = -731.971//9.81= -74.61 sec



The first soln of 4.686 sec is correct as the other one implies that the stone reaches the water before it is even dropped!!



tw = 4.686 s

s = 0.5 gtw^2

s = 0.5 x 9.81 x (4.686)^2

s = 4.905 x 21.958 (m)

s = 107.7069 m

s = 107.7 m



The depth of the well is 107.7 m

Answer 4

V = AT = 9.8 * 5 = 49 m/s

Average V = 49 / 2 = 24.5

Distance = VT = 24.5 * 5 = 122.5 meters