Question

Here's my question; A rock is dropped into a deep hole and 1.6 s later the sound of striking the bottom is heard. If the speed of sound is 340 m/s, what is the depth of the hole?

Answer 1

I tried using V^2=Vnaut^2+2a(X-Xnaut) and it did not a thing for this. I believe this may need to be solved in two parts, but I'm still not entirely sure.

Answer 2

My brain is melting.. No lie.

Answer 3

find the time taken for the stone to go down using second equation of motion in terms of distnance

time taken by sound to travel back = distance/ 340

add both

equals 1.6

Answer 4

X - Xo = .5(Vo + V)t <--- This one?

Answer 5

Because the answer was unfortunately not 1.6 meters...

Answer 6

Although, I did the equation through, and I still am having some difficulty understanding why it wants to melt my face with such fervor.

Answer 7

not that one

s = 1/2 g t^2

this one

Answer 8

find t from this and add it with the other t

Answer 9

The distance traveled by the sound and the rock will be the same. You'll need two equations, both solved for s, and then set equal. So, \[s=at+\frac{ 1 }{ 2 }at^2\] for the rock, and \[s=vt\] for the sound (no acceleration) giving us: \[v_st = at+\frac{1}{2}at^2\].

I think that'll get you there.

Answer 10

Thank you @eighthourlunch, this was extremely helpful and verifiably correct.

Answer 11

You're very welcome, glad I could help.