How Deep is the Hole if a Caveman Hears a Rock Hit the Bottom After 16 Seconds?

[Tauskela]

Homework Statement

A caveman drops a rock from ground-level, into a hole. He counts in seconds until he hears the rock hit the bottom of the hole, and finds that the hole is '16 seconds deep.' Assuming there is no air friction, and the initial velocity of the rock is 0m/s, how deep is the hole? Take into account the speed of sound.

Homework Equations

h = ½g(T1)²
h = v(T2)
T1 + T2 = 16 sec

The Attempt at a Solution

Attempt 1:
h = ½g(T1)²
h=½(9.8m/s²)(16s)²
h=4.9m/s²(256s)
h=1254.4m
h=v(T2)
1254.4m=340.29m/s(T2)
1254.4m/340.29m/s=T2
T2=3.6862s
h=v(T2)
1254.4m=v(3.69s)
1254.4m/3.6862s=v
v=340.29m/s
T1+T2=16 sec
16s+3.6862s = 19.6862s?

I am confused because the two values do not add up...and than how do i know which variable actually represents the depth of the hole...

 

[LowlyPion]

No. The times add to 16 by the statement of the problem.

h = ½g(T1)² = v(T2)

and

T1 + T2 = 16

 

[Tauskela]

Okay, so...
h = ½g(T1)² = v(T2)

Therefore...
h=½(9.8m/s²)(16s)²=v(T2)
h=4.9m/s²(256s)
h=1254.4m=340.29m/s(3.6862s)
h=1254.4m=1254.4m

But, I don't understand, because the answer can't be 1254.4m...

 

[LowlyPion]

Okay, so...
h = ½g(T1)² = v(T2)

Therefore...
h=½(9.8m/s²)(16s)²=v(T2)
h=4.9m/s²(256s)
h=1254.4m=340.29m/s(3.6862s)
h=1254.4m=1254.4m

But, I don't understand, because the answer can't be 1254.4m...

You need to stop putting t1 = 16 into your equation.

The total time is 16.

You have 2 unknown time variables.

T1 + T 2 = 16

Solve the 2 equations for the 2 variables with that in mind.

 

[Tauskela]

Okay, so if I leave T1 as you've said...than that means I have two unknown variables...so how would that even work
h = ½g(T1)²

 

[LowlyPion]

Okay, so if I leave T1 as you've said...than that means I have two unknown variables...so how would that even work
h = ½g(T1)²

Eliminate h first of all, it's the same in both directions isn't it?

½g(T1)² = v(T2)

and

T1 + T2 = 16

Solve.

 

[Tauskela]

Okay, so...I put what you said

½g(T1)² = v(T2)

and

T1 + T2 = 16

And than I sub in the values...

½(9.8m/s²)(T1)²=340.29m/s(T2)
4.9m/s²(T1)²=340.29m/s(T2)
4.9m/s²/340.29m/s=(T2)/(T1)²
0.014399s=(T2)/(T1)²
T1+T2=16

Than how would I proceed from here, am I supposed to combine the two equations in some way?

 

[LowlyPion]

You haven't covered substituting variables in your coursework yet?

T1 = 16 - T2

½g(T1)² = v(T2)

½g(16 - T2)² = v(T2)

It could be a long semester.

 

[Tauskela]

Okay, It's not that I can't substitute, its just that there are two variables, so how am I supposed to solve, I'm really confused because if..


T1 = 16 - T2

...How do I even figure out what one of those variables are in the first place?

Should I assume that T2 = 3.6862s, because 1254.4m/340.29m/s=T2

So than, 16-3.6862 = 12.3138, so is this my answer?

The hole is 12.3138m deep?

But how can this even be, because if I'm not assuming that T1 is even 16s, that messes up my entire equation...

 

[LowlyPion]

...How do I even figure out what one of those variables are in the first place?

Simply evaluate the quadratic:

½g(16 - T2)² = v(T2)

Everything is known but T2

Quadratic Formula maybe?

 

[Tauskela]

½g(16 - T2)² = v(T2)
Okay, I get this far..
4.9m/s²(16-T2)² =340.29m/s

So, the quadratic equation means [-b+-sqrt(b²-4ac)]/2a...

Should I assume that 4.9m/s² = c, 340.29m/s = b, and 16 = a? How do I even work with that 16 because its in the bracket with the T2.

 

[LowlyPion]

You will need to expand it of course.

 

[Tauskela]

Oh right. Expanding.

So (16-T2)²
=16²-T2²?

So is 16² going to be my 'a' value?

[-340.29m/s+-sqrt(340.29m/s²-4(16²)(4.9m/s²))]/2(16²)

Would that be my equation?

 

[LowlyPion]

Oh right. Expanding.

So (16-T2)²
=16²-T2²?

So is 16² going to be my 'a' value?

No.

(16 -t)(16 - t) is not 16²-t²

 

[Tauskela]

(16 -t)(16 - t).

Okay, F.O.I.L rules, so..

96-16t-16t-t²
=96-t², because the other values cancel out?

 

[Tauskela]

Oh, wait. I THINK I figured it out, can you confirm?

t = d/343 + √(2d/g) = 16
solve for d

d + 343√(2d/(9.8)) = 5488
d + 155√d = 5488

let x² = d
x² + 155x – 5488 = 0
quadratic equation:
to solve ax² + by + c = 0
x = [–b ±√(b²–4ac)] / 2a
x = [–155 ±√(24025+21952)] / 2
x = [–155 ± 214] / 2
skipping the negative answer..
x = 30

√d for x
√d = 30
d = 900 meters

 

[LowlyPion]

(16 -t)(16 - t).

Okay, F.O.I.L rules, so..

96-16t-16t-t²
=96-t², because the other values cancel out?

No.

256 - 32t + t²

That equals 343/4.9 * t = 70t = 256 - 32t + t

t² - 102t + 256 = 0

T2 = 2.58
x = 343 * 2.58

884 = 1/2*g*t² ===> T1 = 13.43

T1 + T2 = 2.58 + 13.43 ≈ 16