Question

We drop a rock into the well and we hear a splash after 3 seconds. How deep is the well?

(we hear the splash when the rock hits the water)

Answer 1

Answer:

A stone is dropped into a well. The splash is heard 3.00 seconds later. What is the depth of the well?

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Steve Brown

Answered 2 years ago

A stone is dropped into a well. The splash is heard 3.00 seconds later. What is the depth of the well?

Answer: h≈40.6819m≈133.4708005ft .

I've assumed that by the depth of the well the OP means the distance from the ground to the water in the well. As well as wind resistance, etc, I've also neglected the brain's processing time.

We'll let g be the acceleration of gravity and c be the speed of sound. Also, we'll let t1 be the time it takes for the rock to hit the water and t2 be the time for the sound to travel up to ground level. From the problem, we know that t1+t2=3 seconds.

Normally g is specified with a negative sign since gravity works downward, but in this case, we'll let g be positive so that the 0 height is at ground level and the positive height is the distance to the water below.

The equation for acceleration is simple: a(t)=g .

By integrating the acceleration equation twice with respect to t, using the fact that the initial height is 0, and assuming that the initial velocity is 0, we get the equation for the height.

h(t1)=12gt21

The equation for the sound to travel back to ground level is simply

h(t2)=ct2 ,

but since we know that t1+t2=3 , this gives us

h(3−t1)=c(3−t1) .

We know that both heights are the same, so

12gt21=c(3−t1) .

Solving for t1 we get

t1=−c+c2+6cg√g .

So,

h(t1=−c+c2+6cg√g)=12g(−c+c2+6cg√g)2=cg(c+3g−c2+6cg−−−−−−−√) .

For confirmation, we know that t2=3−t1 and

h(t2=3−t1)=h(t2=3−−c+c2+6cg√g)=cg(c+3g−c2+6cg−−−−−−−√) .

Letting c =343 m/s and g =9.8 m/s 2 ,

h≈40.6819m≈133.4708005ft .