I'm not getting this homework problem correct. Can anyone help me out?

Question

Sound travels at 1100 ft./sec. in still air. Anne finds an old well on a farm she has just bought and wants to know how deep it is. She drops a rock from ground level and hears it splash 7.5 seconds later. Correct to the nearest whole foot, how deep is the well?

Answer 1

The problem can be modeled as a set of simultaneous equations.

d is the depth of the well
tf is the time of the fall
tr is the return of the sound

d=1/2*g*tf^2
d=1100*tr
tf+tr=7.5

so
tr=(7.5-tf)

plug into the second
d=1100*(7.5-tf)

subtract
d-d=1/2*g*tf^2-1100*(7.5-tf)

solve for tf and compute d

This gives a quadratic equation with a positive root and a negative root. The positive root is relevant.

0=16*tf^2+1100*tf-8250

I get that
tf=6.823sec
tr=.677sec
d=745 ft



j

Answer 2

When the rock hits the bottom of the well it creates a sound. This sound moves at 1100 ft./sec so do this
1100/1 = x/7.5
x=1100*7.5
x=8250
the well is 8250 feet deep

Answer 3

Forget the speed of sound. You need to calculate how far the rock will fall in 7.5 seconds. The rock will accelerate for approximately 3 seconds before it reaches terminal velocity (see rate of acceleration at one G). Terminal velocity is around 120 mph. The speed of sound is a negligable factor in this equation. Hope this helps.

Answer 4

that's tough, you'd probably need to know the weight before you knew how long the rock took to reach the bottom, after you knew that, you could figure out how far it is by using the remaining time with the speed of sound. That's how I'd do it anyway.

Or maybe you could just use the acceleration of something when pulled downwards by gravity (it stops accellerating after a certain speed has been reached, I think) but I can't remember it, or the time it takes to reach that speed (it depends on weight anyway, probably), but I hope this points you in the right direction.

Answer 5

I think it is
1100x7.5
and put the nearest whole no