How deep is the well using the speed of sound?

[starji]

how deep is the well? (using speed of sound)

Homework Statement

A boy drops a penny into a deep well. if he hears the splash 5.0 seconds after he releases the penny, how deep is the well? (assume the well is 20degrees C) (also assume there is no air resistance)

Homework Equations

speed of sound: v=331+0.59T and basic kinematics

The Attempt at a Solution

i've found the speed of sound to be 342.8m/s

if t=d/v then the time it takes the sound of the splash to reach the boy's ears is t=d/342.8

the time it takes for the penny to drop into the well is: d=vit+0.5at^2; the vi is 0 so d=4.9t^2 therefore the time it take the penny to drop is t=sqrt(d/4.9)

so from here i should be able to find the distance by 5=d/342.8+ sqrt(d/4.9), but it isn't working and I am getting tiny numbers

 

[gneill]

Your method looks okay. Perhaps you could show the steps you've taken to solve your last equation for d.

 

[starji]

oh ok sry here it is: d=vit+0.5at^2; d=0+0.5(9.8)t^2; d=4.9t^2

 

[leokote]

So far, so good. Now make the substitution: x = sqrt(d), and use the quadratic formula to find x. Then square x to get d.

 

[starji]

wait what? i don't really understand where x is coming from or how that would work since i get x^2=4.9t^2 leaving me with 2 variables and no way to use the quad formula

 

[leokote]

Sorry, I was referring to the last equation you presented in your first post. The working up to that point is correct: now you need to solve that equation for d. I suggest making the substitution x = sqrt(d).

 

[starji]

oh ok good call thanks i'll try it