Solving for Distance of Pit: 2 Seconds & 332m/s

[andorei]

Homework Statement

A boy dropped a stone at a pit. A sound was heard 2 seconds later. What is the distance of the bottom of the pit?

The Attempt at a Solution

Given:
Time - 2 Seconds
Vsound - 332m/s

332m/s * 2s = 664meters.

I think it is somewhat wrong. Maybe purely wrong. I asked my teacher if the question has anything to do with vertical acceleration but my teacher said no.

 

[Doc Al]

You need to consider the time it took for the stone to fall to the bottom of the pit.

 

[andorei]

Only the time is known.

But using (at^2)/2, the answer is reasonable.

 

[andorei]

*Also assuming that the 2seconds is the time it took for the stone to fall to the bottom of the pit.

 

[zj8651731]

2 seconds includes two parts,first,the time sound requires to travel a distance of pit,second,the time stone requires to fall to the bottom

 

[andorei]

@zj8651731 - Does this mean that the question itself is wrong?

 

[Doc Al]

There's nothing wrong with the question. Call the unknown distance to the bottom of the pit X. How long does it take to for the stone to fall that distance? How long does it take for the sound of the stone hitting bottom take to get back to the top? Those two times must add to what total time?