1. ## Word problem

I don't even know how to begin this.

You drop a rock into a deep well. You can't see the rock's impact at the bottom, but you hear it after 7 seconds. The depth of the well is _____ feet. Ignore air resistance. The time that passes after you drop the rock has two components: the time it takes the rock to reach the bottom of the well, and the time that it takes the sound of the impact to travel back to you. Assume the speed of sound is 1100 feet per second.
Note: After seconds the rock has reached a depth of feet where d=16t^2

2. Originally Posted by md56
I don't even know how to begin this.

You drop a rock into a deep well. You can't see the rock's impact at the bottom, but you hear it after 7 seconds. The depth of the well is _____ feet. Ignore air resistance. The time that passes after you drop the rock has two components: the time it takes the rock to reach the bottom of the well, and the time that it takes the sound of the impact to travel back to you. Assume the speed of sound is 1100 feet per second.
Note: After seconds the rock has reached a depth of feet where d=16t^2
$\displaystyle t_1$ = time down

$\displaystyle t_2$ = time for the splash sound to return

* $\displaystyle t_1 + t_2 = 7$

$\displaystyle d = 16t_1^2$

$\displaystyle d = 1100t_2$

* $\displaystyle 16t_1^2 = 1100t_2$

solve the system (*) for $\displaystyle t_1$ or $\displaystyle t_2$ ... then find $\displaystyle d$

3. 2/1 following 16t an exponent?

4. Originally Posted by md56
2/1 following 16t an exponent?
$\displaystyle 16(t_1)^2$ ... the 1 is a subscipt.

5. now I am lost.

6. Originally Posted by md56
now I am lost.
my typo ... should be $\displaystyle 16(t_1)^2$ ... put the opening parenthesis too soon.

7. I don't know what is required . . . let's find SOME of the unknown.

I will use the data above,

T + t = 7 - - - - (1)

Let T = time going down, in seconds

t = time going up, in seconds

d = depth of the well, feet

d = (acceleration of the rock)(Time going down)^2 = 16T^2 - - - - (2)

also, d = (speed of sound)(time going up) = 1100t - - - - (3)

equate (2) and (3)

16T^2 = 1100t

t = 16T^2/1100 - - - - (4)

substitute (4) in (1),

T + t = 7, that is t = 16T^2/1100

T + 16T^2/1100 = 7, cross-multiply

1100T + 16T^2 = 7700,rearrange them

16T^2 + 1100T - 7700 = 0,

it is now quadratic in T, can you solve it now?

8. I am getting 6.403, and it is incorrect

9. Originally Posted by md56
I am getting 6.403, and it is incorrect
6.403 is the time it takes for the rock to get to the bottom after being dropped.

10. T = (5/8)(3(sqrt 473) -55) = 6.4 sec,

T + t = 7,

t = 7 - 6.4 = 0.6 sec.

depth of the well = d = 1100t = 1100(0.6) = 660 feet.

Wow! Is not that too deep for a well?